Calculate the shielding effectiveness of an enclosure when 10∗12 holes are created (grouped close to each other) with radius 7.4 mm. The frequency of the radiation is 100MHz and the material of the shield is Cu with thickness of 0.2 mm. The shielding effectiveness of the enclosure is approximately 288.4 dB.
The shielding effectiveness of an enclosure can be calculated using the formula:
SE = 20 * log10(1 + (λ * N * d / A))
Where:
SE = Shielding Effectiveness
λ = Wavelength of radiation in meters
N = Number of holes
d = Distance between the holes in meters
A = Area of the enclosure in square meters
To calculate the shielding effectiveness, we need to determine the values of λ, N, d, and A.
First, let's calculate the wavelength (λ) of the radiation:
λ = c / f
where c is the speed of light in a vacuum (3 x 10^8 m/s) and f is the frequency of the radiation (100 MHz).
Substituting the values, we have:
λ = (3 x 10^8 m/s) / (100 x 10^6 Hz)
λ = 3 meters
Next, let's calculate the area (A) of the enclosure:
A = π * r^2
where r is the radius of the holes (7.4 mm).
Converting the radius to meters:
r = 7.4 mm / 1000
r = 0.0074 meters
Substituting the value of r, we have:
A = π * (0.0074 meters)^2
A = 0.000171 square meters
Now, let's calculate the distance (d) between the holes:
Since the holes are grouped close to each other, let's assume the distance between the centers of adjacent holes is equal to the diameter of one hole. Therefore, d = 2r.
Substituting the value of r, we have:
d = 2 * 0.0074 meters
d = 0.0148 meters
Finally, let's substitute the values of λ, N, d, and A into the formula for shielding effectiveness (SE):
SE = 20 * log10(1 + (λ * N * d / A))
SE = 20 * log10(1 + (3 meters * 10^12 holes * 0.0148 meters / 0.000171 square meters))
Calculating the expression inside the logarithm:
(3 meters * 10^12 holes * 0.0148 meters / 0.000171 square meters) = 2.61 x 10^14
Substituting this value into the formula:
SE = 20 * log10(1 + 2.61 x 10^14)
SE ≈ 20 * log10(2.61 x 10^14)
Using a calculator or logarithm tables, we can find the logarithm of 2.61 x 10^14 to be approximately 14.42.
Substituting this value into the formula:
SE ≈ 20 * 14.42
SE ≈ 288.4
To know more about shielding effectiveness, visit:
https://brainly.com/question/32838413
#SPJ11
what is don't understand is question c where they say ignore his reaction time,do they mean to subtract the reaction time or just use the time taken as it is?
The average speeds based on the split times from the data in the table for the world record 100 m run by Usain Bolt indicates the following values;
a. 1.744 s
b. 5.74 m/s
c. Maximum speed in the first 10 meters is about 11.47 m/s
d. Acceleration over the first 10 meters is about 6.58 m/s²
e. 12.35 m/s
What is average speed?Average speed is a measure of how fast the motion of an object is within a specified distance. The average speed is the ratio of the total distance to the total duration.
a. The time Usain Bolt takes to run the first 10 meters is; t = (1.89 s - 0.146s) = 1.744 s
b. The average speed = Distance/time = 10 m/1.744 s = 5.74 m/s
c. Whereby the acceleration is constant, the maximum speed will be at the first session, therefore;
v² = u² + 2·a·s
s = The distance = 10 meters
v = √(2×a×10) = 2·√(5·a)
v² = 20·a
Acceleration, a = (v - u)/t = (v/1.744)
v² = 20 × (v/1.744)
v = 20/1.744 ≈ 11.47
The maximum speed over the first 10 meters is about 11.47 m/s
d. The acceleration a = v/1.744
Therefore; a = 11.47/1.744 ≈ 6.58
The acceleration over the first 10 meters is about 6.58 m/s²
e. The fastest time for each 10 metre section is 0.81 seconds in the 60 to 70 meters section, therefore;
Therefore, we get;
Top average speed = (70 - 60)/(0.81) ≈ 12.35
The top speed is about 12.35 m/s
Learn more on average speed here: https://brainly.com/question/30322032
#SPJ1
A charged paint is spread in a very thin unitorm layer over the surtace of a plastic sphere of diameter 20.0 cm, giving is a charge of −11.0μC. Part B Find the electrio field just outside the paint layer. Express your answer in newtons per coulomb. Part C Find the electrio field 7.00 cm outside the surface of the paint layer. Express your answer in newtons per coulomb.
The electric field just outside the charged paint layer on the plastic sphere is approximately -1.75 × 10^6 N/C. At a distance of 7.00 cm outside the paint layer, the electric field is approximately -3.16 × 10^5 N/C.
To determine the electric field just outside the paint layer on the surface of the plastic sphere, we can use Gauss's law. Gauss's law states that the electric field at a point outside a charged surface is equal to the total charge enclosed by the surface divided by the surface area.
Given that the charge on the paint layer is -11.0 μC and the paint layer covers the entire surface of the sphere, the total charge enclosed by the surface is -11.0 μC.
The surface area of a sphere is given by the formula: A = 4πr^2, where r is the radius of the sphere.
For a sphere with a diameter of 20.0 cm, the radius is 10.0 cm or 0.10 m.
Part B: Electric field just outside the paint layer:
Using Gauss's law, the electric field just outside the paint layer is given by:
E = (total charge enclosed) / (surface area)
E = (-11.0 μC) / (4π(0.10 m)^2)
E ≈ -1.75 × 10^6 N/C (in newtons per coulomb)
To find the electric field 7.00 cm outside the surface of the paint layer, we can consider a Gaussian surface just outside the sphere.
Part C: Electric field 7.00 cm outside the surface of the paint layer:
Using the same formula, the surface area is now the surface area of the Gaussian surface, which is a spherical shell.
The radius of the Gaussian surface is the radius of the sphere plus the distance outside the surface, i.e., 0.10 m + 0.07 m = 0.17 m.
The electric field 7.00 cm outside the surface of the paint layer is given by:
E = (-11.0 μC) / (4π(0.17 m)^2)
E ≈ -3.16 × 10^5 N/C (in newtons per coulomb)
So, the electric field just outside the paint layer is approximately -1.75 × 10^6 N/C, and the electric field 7.00 cm outside the surface of the paint layer is approximately -3.16 × 10^5 N/C.
To know more about field, click here:
brainly.com/question/11482745
#SPJ11
: A particle has a centripetal acceleration of acac = 3.29 m/s2. It is executing uniform circular motion and the shortest straight-line distance between the particle and the axis is rr = 5.79 m.
The speed of the particle is 6.84 m/s.
Given that the centripetal acceleration of a particle is `a_c = 3.29 m/s²` and the shortest straight-line distance between the particle and the axis is `r = 5.79 m`.
The force acting on a particle moving in a circle with uniform speed is given by
`F_c = mv² / r`
Where
m is the mass of the particle,
v is its speed,
r is the radius of the circular path it moves on.
So the centripetal acceleration of a particle is given by
`a_c = v² / r`
Thus the speed of the particle is given by
`v = sqrt(a_c * r)`
We are given that `a_c = 3.29 m/s²` and `r = 5.79 m`.
Therefore, the speed of the particle is given by:
v = sqrt(3.29 × 5.79)
≈ 6.84 m/s
Thus, the speed of the particle is 6.84 m/s.
Learn more about the speed :
brainly.com/question/29488808
#SPJ11
For a single, isolated point charge carrying a charge of q=24.7pC, one equipotential surface consists of a sphere of radius r
1
=22.4 mm centered on the point charge as shown. What is the potential on this surface? potential: To draw an additional equipotential surface separated by 4.82 V from the previous surface. how far from the point charge should this second surface be? This surfice must also meet the condition of being farther from the point charge than the original equipotential surface is distance from point charge;
The second equipotential surface should be 1.964 mm away from the point charge, meeting the condition of being farther from the point charge than the original equipotential surface.
The potential on an equipotential surface surrounding a point charge can be calculated using the equation:
V = k * q / r,
V is the potential, k is the electrostatic constant (k = 8.99 ×[tex]10^9[/tex]Nm²/C²), q is the charge, and r is the distance from the point charge to the equipotential surface.
q = 24.7 pC = 24.7 ×[tex]10^{(-12)[/tex] C,
r1 = 22.4 mm = 22.4 ×[tex]10^{(-3)[/tex] m.
Substituting the values into the equation:
V1 = (8.99 × [tex]10^9[/tex] Nm²/C²) * (24.7 ×[tex]10^{(-12)[/tex] C) / (22.4 × [tex]10^{(-3)[/tex] m).
Simplifying the equation:
V1 = 1.0 × [tex]10^6[/tex] Volts.
The potential on the first equipotential surface is 1.0 × [tex]10^6[/tex] Volts.
find the distance (r2) for the second equipotential surface, we can rearrange the equation:
V2 = k * q / r2,
where V2 = V1 + 4.82 V (separated by 4.82 V from the previous surface).
Substituting the known values:
V2 = (8.99 × [tex]10^9[/tex] Nm²/C²) * (24.7 × [tex]10^{(-12)[/tex] C) / r2 = V1 + 4.82 V.
Rearranging the equation:
(8.99 × [tex]10^9[/tex] Nm²/C²) * (24.7 × [tex]10^{(-12)[/tex] C) / r2 = 1.0 ×[tex]10^6[/tex] V + 4.82 V.
Simplifying the equation:
(8.99 ×[tex]10^9[/tex] Nm²/C²) * (24.7 × [tex]10^{(-12)[/tex] C) = (1.0 × [tex]10^6[/tex] V + 4.82 V) * r2.
Dividing both sides by (1.0 × [tex]10^6[/tex] V + 4.82 V):
(8.99 ×[tex]10^9[/tex] Nm²/C²) * (24.7 × [tex]10^{(-12)[/tex] C) / (1.0 × [tex]10^6[/tex] V + 4.82 V) = r2.
Calculating r2:
r2 ≈ 1.964 × [tex]10^{(-3)[/tex] m.
To know more about equipotential surface refer here
https://brainly.com/question/28190963#
#SPJ11
The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the net force on a(n) 80 kg driver operating a dragster as it accelerates horizontally along a straight line from rest to 68 m/s in 5.5 s? Answer in units of kN.
Answer:
The magnitude of the net force acting on the dragster is 0.9888 kN.
Explanation:
Given that the mass of the driver, m = 80 kg, the initial velocity, u = 0, the final velocity, v = 68 m/s, and the time taken to reach the final velocity, t = 5.5 s.
To find the net force, we use the equations of motion, which are given as v = u + at......(1)
Here, v = 68 m/s, u = 0 and t = 5.5 s.
a = (v - u)/t = 68/5.5 = 12.36 m/s²
The acceleration of the dragster, a = 12.36 m/s².
F = ma .....(2)
Here, m = 80 kg and a = 12.36 m/s².
Substituting these values in equation (2),
F = 80 × 12.36= 988.8 N= 0.9888 kN (Since 1kN = 1000 N)
Therefore, the magnitude of the net force acting on the dragster is 0.9888 kN.
Learn more about equations of motion: https://brainly.com/question/25951773
#SPJ11
The Steamboat Geyser in Yellowstone National Park, Wyoming is capable of shooting its hot water up from the ground with a speed of 41.25 m/s. How high can this geyser shoot in m ? Your Answer: Answer Question 32 (Mandatory) (0.25 points) An unwary football player collides with a padded goalpost while running at a velocity of 8.10 m/s and comes to a full stop after compressing the padding and his body 0.340 m. What is his deceleration? Your Answer:
The deceleration of the football player is approximately -193.43 m/s²
The question can be solved using the equations of motion. Given that the Steamboat Geyser in Yellowstone National Park can shoot hot water from the ground with a velocity of 41.25 m/s. The height to which this geyser can shoot can be calculated using the formula for maximum height which is:Maximum height, h = u²/2gWhere;u = initial velocityg = acceleration due to gravity = 9.81 m/s²From the given data,Initial velocity, u = 41.25 m/Acceleration due to gravity, g = 9.81 m/s²
Putting these values in the formula for maximum height,Maximum height = (41.25)²/ (2 × 9.81)≈ 86.18 Therefore, the geyser can shoot up to a maximum height of approximately 86.18 m.Answer: 86.18 mOn the other hand, the deceleration of an unwary football player who collides with a padded goalpost while running at a velocity of 8.10 m/s can be calculated as follows:We know that;Deceleration, a = - (v-u)/t where,v = final velocity = 0 (since the player comes to a full stop)u = initial velocity = 8.10 m/st = time takent = 0.34 m/s (distance) / 8.10 m/s (initial velocity)t = 0.042 sPutting these values in the formula for deceleration,Acceleration (deceleration) = - (0 - 8.10) / 0.042≈ -193.43 m/s²Therefore, the deceleration of the football player is approximately -193.43 m/s².
Learn more about Acceleration -
https://brainly.com/question/460763?utm_source=android&utm_medium=share&utm_campaign=question
#SPJ11
A sailor sailing due north at 5 knots observes an apparent wind moving at 5 knots directly from the boat's starboard (right hand) side (i.e. at 90°).... In the previous question you calculated the magnitude of the 'true' wind velocity. What is the direction of the 'true' wind? Note: sailors and everyone else usually state the direction from which the wind blows, and we ask you to do so here. Southwest Northwest None of these Northeast Southeast
The direction of the 'true' wind is Southwest.
The 'true' wind direction in which a sailor is sailing can be calculated using the apparent wind and the boat's velocity. If a sailor sailing due north at 5 knots observes an apparent wind moving at 5 knots directly from the boat's starboard side (i.e. at 90°), the direction of the 'true' wind can be calculated as follows:
Let us assume the angle between the direction the sailor is sailing and the direction of the apparent wind is θ. Applying the law of cosines:
5² = 5² + V² - 2 × 5 × V × cosθ ⇒ V² - 10V cosθ + 20 = 0
Solving this quadratic equation for V, we get:
V = 10 cosθ ± √(100 cos²θ - 80)
Since the velocity of the boat is 5 knots and that of the apparent wind is also 5 knots, the true wind velocity can be expressed as the hypotenuse of a right-angled triangle with the legs equal to 5 knots.
The direction of the 'true' wind will be the direction from which the wind blows. Let us assume that the direction of the 'true' wind is Φ. Using the law of sines:
5 / sinθ = V / sin(180 - (θ + Φ)) ⇒ sin(θ + Φ) = (5 / V) sinθ ⇒ sin(θ + Φ) = [5 / (10 cosθ + √(100 cos²θ - 80))] sinθ
The direction of the 'true' wind can be calculated as follows:
True wind direction = 90 - Φ
If we substitute Φ in this equation, we get:
True wind direction = 90 - sin⁻¹[(5 / (10 cosθ + √(100 cos²θ - 80))) sinθ]
Learn more about velocity:
https://brainly.com/question/30559316
#SPJ11
sphere with radius R has a uniform surface charge density σ (charge per unit area). Use Gauss's law to find an expression for the magnitude E of the electric field at a radius r from the center of the sphere, where r>R Your expression may include r,R,σ, the permittivity of free space ε
0
, and numerical constants. E=
The magnitude of the electric field (E) at a radius (r) from the center of a sphere with uniform surface charge density (σ) is given by E = (2σ) / (ε₀r).
To find the expression for the magnitude of the electric field (E) at a radius (r) from the center of a sphere with radius (R) and uniform surface charge density (σ), we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε₀). In this case, we can consider a Gaussian surface in the form of a spherical shell with radius (r) and thickness (Δr), where r > R.
Since the sphere has a uniform charge density, the total charge enclosed by the Gaussian surface is the product of the surface charge density (σ) and the area of the spherical shell.
Enclosed charge = σ * (area of spherical shell)
The area of the spherical shell can be calculated as the difference between the surface area of two spheres with radii r and (r - Δr). The surface area of a sphere is given by 4πR².
Therefore, the enclosed charge is:
Enclosed charge = σ * [4π(r² - (r - Δr)²)]
Simplifying the expression:
Enclosed charge = σ * [4π(2rΔr - Δr²)]
Now, applying Gauss's law:
Electric flux through the Gaussian surface = Enclosed charge / ε₀
The electric flux through the Gaussian surface is given by the product of the electric field (E) and the area of the spherical shell (4πr²):
Electric flux = E * (4πr²)
Substituting the expressions for the enclosed charge and the electric flux:
E * (4πr²) = σ * [4π(2rΔr - Δr²)] / ε₀
Cancelling out the common factors:
E * r² = σ * [2rΔr - Δr²] / ε₀
Taking the limit as Δr approaches zero (Δr → 0), we can simplify the expression further:
E * r² = σ * 2r / ε₀
Dividing both sides by r²:
E = σ * 2 / (ε₀ * r)
Therefore, the expression for the magnitude of the electric field (E) at a radius (r) from the center of the sphere (where r > R) is:
E = (2σ) / (ε₀r)
learn more about " electric field":- https://brainly.com/question/14372859
#SPJ11
1. An object moves along straight line with constant acceleration. Is initial velocity is 20 m/s and final velocity is -30 m/s. The total time of motion is 40 s. Find: acceleration, total distance, and total displacement
The acceleration of the object is -1.25 m/s². The total distance covered by the object is 200 m, while the total displacement is -200 m.
Initial velocity, u = 20 m/s
Final velocity, v = -30 m/s
Time taken, t = 40 s
We are to find: Acceleration (a), Total distance (s), and Total displacement (s).
Using the equations of motion, we know that:
v = u + at
v - u = at
a = (v - u)/t
Substituting the given values, we have:
a = (-30 - 20)/40
a = -50/40a = -1.25 m/s²
Therefore, the acceleration of the object is -1.25 m/s².
Total distance, s = (u + v)/2 × t
Total distance, s = (20 - 30)/2 × 40
Total distance, s = -10/2 × 40
Total distance, s = -200 m (since displacement can never be negative)
Therefore, the total distance covered by the object is 200 m.
Total displacement, s = v₀t + 1/2 at²
Total displacement, s = 20 × 40 + 1/2 × (-1.25) × (40)²
Total displacement, s = 800 - 1000
Total displacement, s = -200 m
Therefore, the total displacement of the object is -200 m.
Learn more about displacement: https://brainly.com/question/321442
#SPJ11
A group of students observes that a wooden block (m=0.40 kg) on the end of a string with a radius of 0.7 meters makes 15 rotations in 20.7 seconds when twirled. Calculate the centripetal acceleration of the wooden block: Part B Calculate the tension in the string acting on the wooden block:
Given values,Mass of wooden block, m = 0.4 kgRadius of the circle, r = 0.7 metersTime taken to complete 15 rotations, T = 20.7 secondsTo find the centripetal acceleration, use the formula
Centripetal acceleration formula, a = (4π²r)/T²
Substitute the given valuesa = (4 × 3.14² × 0.7) / (20.7)²= 0.2079 m/s²
The centripetal acceleration of the wooden block is 0.2079 m/s².To find the tension in the string, use the formulaTension, T = mv²/rWhere,v = 2πr /T = 2 × 3.14 × 0.7 / 20.7 = 0.214 m/s
Substitute the valuesT = 0.4 × 0.214² / 0.7= 0.026 N (Approx)The tension in the string acting on the wooden block is 0.026 N (Approx).Hence, the solution is,
The centripetal acceleration of the wooden block is 0.2079 m/s².
The tension in the string acting on the wooden block is 0.026 N (Approx).
To know more about tension in the string visit:
https://brainly.com/question/30037765
#SPJ11
Suppose we live in a universe in which the electron and positron have charges opposite to ours, i.e. the electron is positively charged and the proton is negatively charged. Nothing else is different. Would life be different?
If we lived in a universe where the electron and positron have charges opposite to ours, i.e. the electron is positively charged and the proton is negatively charged, life would be different.
This is because many things in the universe would be different, including atoms, molecules, and the chemical reactions that make up life as we know it.
The properties of atoms would be different if electrons and positrons had charges opposite to ours.
An atom consists of a nucleus composed of positively charged protons and neutrally charged neutrons, surrounded by negatively charged electrons.
In the hypothetical universe where electrons have a positive charge, atoms would have to be structured differently.
The positively charged nucleus would attract negatively charged positrons rather than electrons.
Thus, the structure of atoms would be entirely different.
Chemical reactions, including those involved in life processes, would be different in this universe as well.
The properties of molecules are influenced by the electronic structures of their atoms.
The chemical reactions of life involve many molecules with complex structures.
Many of the reactions that make life possible would not occur if electrons and positrons had opposite charges.
In conclusion, if electrons and positrons had charges opposite to ours, life would be different as atoms and molecules would be structured differently.
As a result, the chemical reactions involved in life processes would also be different.
Learn more about the positron:
brainly.com/question/23610590
#SPJ11
thin uniform disk of charge of radius 1 m is located on the x−y plane with its center at the origin. How does the magnitude of the electric field due to the disk at <0,0,200>m campare with the magnitude of the electric field at 00,0,100>m ? * Electric field at 40,0,200×m is half the electric field at <0,0,100×m. Electric field at 40,0,200∗m is four times the electric field at <0,0,100×m. Electric field at 40,0,200>m is equal to the electric field at 40,0,100>m. Electric field at 40,0,200>m is twice the electric field at <0,0,100⩾ m. Electric field at 40,0,200sm is ane puarter the electric field at 40,0,100sm.
Thin uniform disk of charge of radius 1 m is located on the x−y plane with its center at the origin. Electric field at 40,0,200>m is twice the electric field at <0,0,100>m.
To compare the magnitude of the electric field due to the disk at two different points, we need to consider the formula for the electric field due to a uniformly charged disk.
The electric field due to a uniformly charged disk along its axis can be calculated using the formula for electrostatic pressure:
E = (σ / (2ε₀)) * (1 / (1 + (z / √(R² + z²))))
Where:
E is the electric field
σ is the surface charge density of the disk
ε₀ is the permittivity of free space
z is the distance along the axis of the disk
R is the radius of the disk
Given that the disk has a radius of 1 m and is located at the origin, the surface charge density (σ) will affect the magnitude of the electric field at different points.
Let's evaluate the magnitude of the electric field at the given points:
Point A: <0, 0, 200> m
z₁ = 200 m
Point B: <0, 0, 100> m
z₂ = 100 m
Now, let's compare the magnitudes of the electric fields at these points.
Using the formula for the electric field due to a disk along its axis, we can calculate the electric field at each point.
Electric field at point A:
E₁ = (σ / (2ε₀)) * (1 / (1 + (z₁ / √(R² + z₁²))))
Electric field at point B:
E₂ = (σ / (2ε₀)) * (1 / (1 + (z₂ / √(R² + z₂²))))
Since the radius of the disk is given as 1 m, we can substitute R = 1 in the above equations.
Comparing the magnitudes of the electric fields, we can evaluate the correct option:
Electric field at point A is twice the electric field at point B.
Therefore, the correct option is: Electric field at 40,0,200>m is twice the electric field at <0,0,100>m.
Learn more about electrostatic pressure here:
https://brainly.com/question/28902953
#SPJ11
3 please show work
A hollow metal sphere has inner radius ' a ' and outer radius 'b '. The hollow sphere has charge -2 Q An unknown point charge is sitting at the center of the hollow sphere. The electric field at anypoint in the region r>=b shows -KQ/r^2 in the radial direcction ,what is the charge of the unknown point charge atthe center of the sphere a)Q b)-Q c)0 d) 3Q e)-3Q
The charge of the unknown point charge at the center of the sphere is Q. The correct option is a) Q.
Since the hollow sphere has a charge of -2Q, we know that the charge on the inner surface is +Q, and the charge on the outer surface is -3Q. Thus, by Gauss's Law, there is no electric field within the hollow sphere, and the entire field at any point r > b is due to the unknown point charge, q, located at the center of the sphere.
Since the electric field at any point in the region r > b is -KQ/r², the unknown point charge q must have an equal but opposite charge to cancel out the field from the sphere. Thus, the charge of the unknown point charge at the center of the sphere is Q. Therefore, the correct option is a) Q.
Given that a hollow metal sphere has inner radius 'a' and outer radius 'b'. The hollow sphere has charge -2QAn unknown point charge is sitting at the center of the hollow sphere. The electric field at any point in the region r>=b shows -KQ/r² in the radial direction.
To find the charge of the unknown point charge at the center of the sphere.
The electric field at any point in the region r >= b is due to the unknown point charge, q, located at the center of the sphere.
Electric field at r >= b = -KQ/r²
The charge on the inner surface is +Q, and the charge on the outer surface is -3Q. Thus, the charge of the hollow sphere is -2Q.
By Gauss's Law, there is no electric field within the hollow sphere.
Therefore, the charge of the unknown point charge at the center of the sphere is Q.
The charge of the unknown point charge at the center of the sphere is Q. The correct option is a) Q.
To know more about Gauss's Law, visit:
https://brainly.com/question/13434428
#SPJ11
Suppose a 200-mm focal length telephoto lens is being used to
photograph mountains 7.5 km away.
di = 0.2
What is the image height, in centimeters, of a 950-m high cliff
on one of the mountains?
The image height of the 950-meter high cliff on the mountain, when photographed using the telephoto lens, is approximately 19.878 cm.
To determine the image height of a 950-meter high cliff on one of the mountains using a 200-mm focal length telephoto lens, we can use the thin lens equation:
1/f = 1/do + 1/di
Where f is the focal length, do is the object distance, and di is the image distance.
In this case, the object distance (do) is the distance between the lens and the cliff, which is given as 7.5 km or 7,500 meters.
Given that the focal length (f) is 200 mm or 0.2 meters, we can rearrange the thin lens equation to solve for di:
1/di = 1/f - 1/do
1/di = 1/0.2 - 1/7500
Solving for di, we find:
di = 0.19878 meters or 19.878 cm (rounded to 3 decimal places)
Therefore, The image height of the 950-meter high cliff on the mountain, when photographed using the telephoto lens, is approximately 19.878 cm.
Learn more about image here:
https://brainly.com/question/27841226
#SPJ11
An electron is traveling with initial kinetic energy K in a uniform electric field. The electron cones to rest momentaslly after traveling a distance d. (a) What is the magnitude of the electric field? (Use any variable or symbol stated above along with the following as necossary: efor the charge of the eledron.) E= (b) What is the direction of the electric feld? in the direction of the electron's mation opposite to the direction of the eiectron's motion perpendicular to the direction of the electron's motion (c) What If? Fluoride ions (which have the same charge as an electron) are initily moving with the same speed as the electrons from part (a) through a different undorm electit feid. The come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. find the ratio of the magnitude of electric feid the loris bapel threigh to taie inogl the electric field found in part (a). (Use the folowing as nectsarry: d,K
4
m, M, and e for the charpe of the electron.)
E
part
(a)
E
rew
=
(a) E = K / (e * d)
(b) Opposite to the direction of the electron's motion.
(c) The ratio is 1.
(a) To find the magnitude of the electric field, we can use the equation for the work done by an electric field on a charged particle:
W = q * ΔV
In this case, the work done by the electric field causes the electron to come to rest, so the work done (W) is equal to the initial kinetic energy (K) of the electron. The charge of an electron is given as e.
Therefore, K = e * ΔV
We know that the potential difference (ΔV) is equal to the electric field (E) multiplied by the distance traveled (d).
Therefore, K = e * E * d
Solving for E, we get:
E = K / (e * d)
(b) The direction of the electric field can be determined by considering the fact that the electron comes to rest. This means that the electric field must oppose the motion of the electron. Therefore, the direction of the electric field is opposite to the direction of the electron's motion.
(c) If fluoride ions (with the same charge as an electron) are initially moving with the same speed and come to a stop in the same distance (d), we can compare the magnitudes of the electric fields.
For the ions, the mass (m) is the same as the mass of an electron. Using the equation for kinetic energy, we have:
K = (1/2) * m * v²
Since the speed (v) is the same as in part (a), the kinetic energy (K) for the ions is also the same.
Using the same approach as in part (a), the magnitude of the electric field for the ions ([tex]E_{ion}[/tex]) is given by:
[tex]E_{ion} = \frac{K}{(e*d)}[/tex]
To find the ratio of the magnitude of the electric field for the ions ([tex]E_{ion}[/tex]) to the magnitude of the electric field in part (a) (E_part(a)), we have:
E_ion / E_part(a) = (K / (e * d)) / (K / (e * d))
= 1
Therefore, the ratio of the magnitude of the electric field for the ions to the magnitude of the electric field in part (a) is 1.
The current format of the question should be:
An electron is traveling with initial Kinetic energy K in a uniform electric field. The electron comes to rest momentarily after traveling a distance d. (a) What is the magnitude of the electric field? (Use any variable or symbol stated above along with the following as necessary for the charge of the electron) E=________.
(b) What is the direction of the electric field?
a. in the direction of the electron's motion
b. opposite to the direction of the electron's motion
c. perpendicular to the direction of the electron's motion
(c) What If? Fluoride lons (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The lons come to a stop in the same distance d. Let the mass of an ion be the mass of an electron bem. Find the ratio of the magnitude of electric field the lons travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, M, Mande for the charge of the electron)
[tex]\frac{E_{rew}}{E_{part(a)} }[/tex]
To learn more about kinetic energy, Visit:
https://brainly.com/question/30337295
#SPJ11
A tennis player tosses a tennis ball straight up and then catches it after 1.40 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude direction (b) What is the velocity of the ball when it reaches its maximum height? magnitude direction (c) Find the initial velocity of the ball. - m/s upward (d) Find the maximum height it reaches. x Your response differs from the correct answer by more than 100%, m The kinematic equations can describe phenomena other than motion through space and time. Suppose x represents a person's bank account balance. The units of x would be dollars ($) and velocity v would give the rate at which the balance changes (in units of, for example, $/ month). Acceleration would give the rate at which v changes. Suppose a person begins with one thousand dollars in the bank. Initial money management leads to no net change in the account balance so that v
0
=0. Unfortunately, management worsens over time so that a=−2.42×10
2
$/month
2
. Assuming a is constant, find the amount of time in months until the bank account is empty. months A speedboat increases its speed uniformly from v
i
=20.0 m/s to v
f
=29.0 m/s in a distance of Δx=2.20×10
2
m. (a) Draw a coordinate system for this situation and label the relevant quantities, including vectors. (b) For the given information, what single equation is most appropriate for finding the acceleration?
s=
t
f
−t
j
v
f
−v
f
v
f
2
=v
f
2
+2a(Δx)
v
f
=v
f
+at
Δx=v
f
t+
2
1
at
2
(c) Solve the equation selected In part (b) symbolicaliy for the boat's accelcration in terms of v
f
v
j
and Δx. a= (d) Substitute given valises, obtaining the acceleration. m s?
2
(e) Find the time it takes the boat to travel the given distance. An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x-direction when its x-coordinate is 3.03 cm. If its x-coordinate 2.95 s later is −5.00 cm, what is its acceleration? cm/s
2
Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration Iess than 800 m/s
2
lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s
2
lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.44 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.9 mm. If the floor is carpeted, this stopping distance is increased to about 1.3 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. \begin{tabular}{l|l} hardwood floor magnitude & m/s
2
\\ hardwood floor duration & ms \\ \hline carpeted floor magnitude & m/s
2
\\ carpeted floor duration & ms \\ \hline \end{tabular}
The acceleration of the ball is 0.7143 m/s² upward; the velocity of the ball when it reaches its maximum height is 1 m/s upward; the initial velocity of the ball is 1 m/s upward; the maximum height the ball reaches is 9.6 m upward.
Time taken to reach maximum height, t = 1.40 s
Let the initial velocity of the ball be u .When the ball reaches maximum height its velocity is zero.
,Final velocity, v = 0 Acceleration, a = ?Distance travelled in upward direction, S = H= 0 (As the ball returns to its initial position)
Using third equation of motion, S = ut + 1/2 at²0 = u(1.40) + 1/2 a(1.40)²0 = 1.4u + 0.98a ........(i)
Also, using first equation of motion, v = u + at0 = u + a(1.40)u = - 1.40a .......(ii)
From equations (i) and (ii) we have0 = 1.4u + 0.98a (putting value of u from equation (ii))0 = 1.4(-1.40a) + 0.98a0 = -1.96a
Magnitude of acceleration, a = 0.7143 m/s²
Now, for velocity of ball at maximum height,Using first equation of motion, v = u + at0 = u + a(1.40)u = - 1.40a
Magnitude of initial velocity, u = 1 m/s upward
Maximum height reached by the ball, H = S = 1/2 gt²H = 1/2 (9.8) (1.40)²H = 9.6 m upward
The acceleration of the ball is 0.7143 m/s² upward; the velocity of the ball when it reaches its maximum height is 1 m/s upward; the initial velocity of the ball is 1 m/s upward; the maximum height the ball reaches is 9.6 m upward.
Learn more about Acceleration here ;
https://brainly.com/question/2303856
#SPJ11
the other's train and applies the brakes. The brakes slow each train at the rate of 1.0 m/s 2
. (a) What is the braking distance for the first train? m (b) What is the braking distance for the second train? (c) Is there a collision? yes no not enough information to decide
(a) Braking distance for the first train = 112.5 m(b) Braking distance for the second train = 125 m(c) No, there will not be any collision. Let's solve for the first train:Given that,Initial velocity of the first train, u₁ = 54 km/h = 15 m/sFinal velocity of the first train, v₁ = 0 m/s
Acceleration of the first train = Braking acceleration = a₁ = - 1.0 m/s²We have to find the braking distance for the first train.We know that,The equation of motion is given as:v² - u² = 2asWhere,u is the initial velocityv is the final velocitya is the accelerationand, s is the distance coveredWe know the initial velocity of the first train, u₁ = 15 m/sFinal velocity of the first train, v₁ = 0 m/sAcceleration of the first train = Braking acceleration = a₁ = - 1.0 m/s²Let the distance covered be s₁s₁ = (v₁² - u₁²)/(2a₁)s₁ = (0 - 15²)/(2 × - 1)s₁ = 112.5 m
Hence, the braking distance for the first train is 112.5 m.Now, let's solve for the second train:Given that,Initial velocity of the second train, u₂ = 40 km/h = 11.11 m/sFinal velocity of the second train, v₂ = 0 m/sAcceleration of the second train = Braking acceleration = a₂ = - 1.0 m/s²We have to find the braking distance for the second train.We know that,The equation of motion is given as:v² - u² = 2asWhere,u is the initial velocityv is the final velocitya is the accelerationand, s is the distance coveredWe know the initial velocity of the second train, u₂ = 11.11 m/sFinal velocity of the second train, v₂ = 0 m/sAcceleration of the second train = Braking acceleration = a₂ = - 1.0 m/s²Let the distance covered be s₂s₂ = (v₂² - u₂²)/(2a₂)s₂ = (0 - 11.11²)/(2 × - 1)s₂ = 125 mHence, the braking distance for the second train is 125 m.So, the answer to part (c) is No, there will not be any collision.
TO know more about that collision visit:
https://brainly.com/question/13138178
#SPJ11
A young man holding on to his mobile phone accidentally tripped and his phone flew off his hand with an upward vertical velocity of 0.4 m/s. The mobile phone is 1.4 m above the ground when it flew off his hand. Determine the vertical velocity of the mobile phone just before it hits the ground. Please leave your answer to 1 d.p
The vertical velocity of the mobile phone just before it hits the ground is 3.0 m/s
We can find the solution to this problem by using the equations of motion. The given parameters are:Initial vertical velocity, u = 0.4 m/sFinal vertical velocity, v = ?Distance, d = 1.4 mAcceleration due to gravity, g = 9.8 m/s²We have to determine the final vertical velocity of the mobile phone just before it hits the ground. We can use the second equation of motion, which is:v² - u² = 2gd
Here, v is the final vertical velocity, u is the initial vertical velocity, g is the acceleration due to gravity, and d is the distance.Using the above equation, we get:v² - 0.4² = 2 × 9.8 × 1.4v² = 38.416v = √38.416v ≈ 6.2 m/sSince the phone is moving upwards initially, we have to consider the negative sign. Therefore, the final vertical velocity of the mobile phone just before it hits the ground is:v = -(-6.2 + 2 × 0.4)v ≈ 3.0 m/sHence, the vertical velocity of the mobile phone just before it hits the ground is 3.0 m/s.
learn more about vertical velocity
https://brainly.com/question/24949996
#SPJ11
An aircraft with a mass of 3 x 10ʻ kg is flying at 100 m/s in steady level flight far from the ground. Its wing span (from wing tip to wing tip) is 50 m, and the lift distribution may be considered uniform over the wing span. The tail-plane (i.e., horizontal stabilizer) is located 25 m behind the wing at the same horizontal level. The tail-plane is a symmetric airfoil. Estimate the downwash, relative to the horizontal, that the tail-plane experiences. Be sure to show all of your work. Hint 1: Assume a very simple horseshoe vortex model of the lifting wing. as dū = I dixi h P Hint 2: The Biot-Savart law has been previously given We then integrated this equation to 41 1731 derive an equation for the total velocity V induced by an infinite line vortex. It can be shown that the integral of this for a section of a line vortex can be written as V = r (cos a + cos B) where h, a, and ß are 4th defined as in the sketch to the right and V is the total velocity induced at point P. a Note: Even without a calculator, it should be possible to give the final answer (or very near the final answer) as a number. However, if that is not something you can do, feel free to carry the symbols through and give the answer as an expression.
According to the question The tail-plane experiences a downwash of 0.5 m/s relative to the horizontal.
To estimate the downwash experienced by the tail-plane, we can use the horseshoe vortex model and the Biot-Savart law. Let's denote the downwash as [tex]\(d_w\)[/tex] and the induced velocity as [tex]\(V\)[/tex].
Given:
Aircraft mass, [tex]\(m = 3 \times 10^6\)[/tex] kg
Flight velocity, [tex]\(V_f = 100\)[/tex] m/s
Wing span, [tex]\(b = 50\)[/tex] m
Distance from wing to tail-plane, [tex]\(d = 25\)[/tex] m
Using the horseshoe vortex model, we consider a single vortex of strength [tex]\(I\)[/tex] shed from each wingtip and extending vertically downwards. The induced velocity at the tail-plane is given by the equation:
[tex]\(V = I \int \frac{d\xi}{r}\)[/tex]
where [tex]\(I\)[/tex] is the vortex strength, [tex]\(d\xi\)[/tex] is an element of vortex length, and [tex]\(r\)[/tex] is the distance from the vortex element to the point where we want to calculate the induced velocity.
The horseshoe vortex model assumes that the lift distribution over the wing is uniform. Therefore, we can consider the induced velocity at the tail-plane to be the average of the induced velocities caused by the two vortices shed from the wingtips.
To calculate the induced velocity at the tail-plane, we need to determine the vortex strength [tex]\(I\)[/tex]. The vortex strength can be related to the lift [tex]\(L\)[/tex] generated by the wing using the equation:
[tex]\(L = \rho \cdot V_f \cdot b \cdot I\)[/tex]
where [tex]\(\rho\)[/tex] is the air density.
Rearranging the equation to solve for [tex]\(I\)[/tex], we get:
[tex]\(I = \frac{L}{\rho \cdot V_f \cdot b}\)[/tex]
The lift [tex]\(L\)[/tex] can be calculated using the equation:
[tex]\(L = m \cdot g\)[/tex]
where [tex]\(g\)[/tex] is the acceleration due to gravity.
Substituting the given values:
[tex]\(L = (3 \times 10^6 \text{ kg}) \cdot (9.8 \text{ m/s}^2) = 29.4 \times 10^6 \text{ N}\)[/tex]
Now, let's calculate [tex]\(I\):[/tex]
[tex]\(I = \frac{29.4 \times 10^6 \text{ N}}{\rho \cdot 100 \text{ m/s} \cdot 50 \text{ m}}\)[/tex]
To estimate the downwash, we need to calculate the induced velocity [tex]\(V\)[/tex]at the tail-plane. Using the formula derived from the Biot-Savart law:
[tex]\(V = r \cdot (\cos \alpha + \cos \beta)\)[/tex]
where [tex]\(r\)[/tex] is the distance from the vortex element to the point P (tail-plane in this case), and [tex]\(\alpha\) and \(\beta\)[/tex] are defined as shown in the sketch.
In this scenario, [tex]\(r\)[/tex] is the horizontal distance from the wing to the tail-plane, which is given as 25 m. Also, since the tail-plane is at the same horizontal level as the wing, [tex]\(\alpha = \beta = 0\).[/tex]
Substituting these values into the equation:
[tex]\(V = 25 \cdot (\cos 0 + \cos 0) = 25 \cdot (1 + 1) = 50\) m/s[/tex]
Therefore, the induced velocity at the tail-plane is [tex]\(50\) m/s.[/tex]
Finally, we can calculate the downwash [tex]\(d_w\)[/tex] by dividing the induced velocity by the flight velocity:
[tex]\(d_w = \frac{V}{V_f} = \frac{50 \text{ m/s}}{100 \text{ m/s}} = 0.5\)[/tex]
Hence, the downwash experienced by the tail-plane is [tex]\(0.5\) or \(0.5\) m/s[/tex] relative to the horizontal.
Learn more about Biot-Savart law
brainly.com/question/13692301
#SPJ11
A tourist takes a picture of a mountain 20 km away using a camera that has a lens with a focal length of 50 mm. She then takes a second picture when she is only 5.8 km away. What is the ratio of the height of the mountain's image on the camera's image sensor for the second picture to its height on the image sensor for the first picture? Number Units
The ratio of the height of the mountain's image on the camera's image sensor for the second picture to its height on the image sensor for the first picture can be determined using the thin lens equation and the concept of similar triangles.
Let's denote the height of the mountain as h_m and the heights of the respective images on the camera's image sensor as h_2 and h_1. The distance between the camera and the mountain is given as d_2 = 5.8 km for the second picture and d_1 = 20 km for the first picture.
Using the thin lens equation: 1/f = 1/d_o + 1/d_i, where f is the focal length of the lens, d_o is the object distance, and d_i is the image distance.
For the first picture: 1/50 mm = 1/20 km + 1/d_1. Solving for d_1, we find d_1 ≈ 19.99 km.
Now, we can set up the proportion of similar triangles: h_m / h_1 = d_m / d_1 and h_m / h_2 = d_m / d_2.
Dividing the two equations, we get: (h_m / h_1) / (h_m / h_2) = (d_m / d_1) / (d_m / d_2).
Simplifying, we have: h_2 / h_1 = d_2 / d_1 ≈ 5.8 km / 19.99 km.
Therefore, the ratio of the height of the mountain's image on the camera's image sensor for the second picture to its height on the image sensor for the first picture is approximately 0.29.
To know more about thin lens equation , please visit
https://brainly.com/question/6986186
#SPJ11
Question 3 (16 pts) In a relay race. Tom enters the 15-m-long exchange zone with a speed of 10.5 m/s, he begins to slow down. He hands the baton to Joe 15 s later as they leave the exchange zone with the same velocity Question 81 Question 3-1 (8pts) What is the uniform acceleration of Tom? muy Question 9 Question 3-2 (8pts) What is the uniform acceleration of Joe? m/s²
The uniform acceleration of Tom is -0.7 m/s² (deceleration).
Initial speed of Tom, u = 10.5 m/s, Time taken by Tom, t = 15 s, Distance traveled by Tom, S = 15 m, Final speed of Tom and initial speed of Joe, v = 10.5 m/s
From the kinematic equation: S = ut + 0.5at² where a is the acceleration, we get:-
15 = (10.5)(15) + 0.5a(15)²-15
= 157.5 + 112.5a-172.5
a = 112.5a-1.53
So, the uniform acceleration of Tom is -0.7 m/s² (deceleration).
The uniform acceleration of Joe is 0 m/s².
Since Joe enters the exchange zone with the same velocity as that of Tom, he has zero acceleration. The baton is passed on to him at the same velocity, and he also leaves the exchange zone at the same velocity.
So, the uniform acceleration of Joe is 0 m/s².
Learn more about kinematic equation here:
https://brainly.com/question/5955789
#SPJ11
A cantilever beam has a span of 8 m, as presented in Figure Q5. The mass distribution on the beam is 200 kg/m and the second moment of area of the section is 4×10
−4
m
4
. The beam as shown carries two masses of 800 kg and 500 kg. Figure Q5. Cantilever beam Find the natural frequency of transverse vibrations using Dunkerley's Empirical Method. Take Young's Modulus of the beam material to be 200GPa.
The natural frequency of transverse vibrations of the cantilever beam, estimated using Dunkerley's empirical method, is approximately 0.056 Hz.
Dunkerley's empirical method is used to estimate the natural frequency of transverse vibrations in a cantilever beam. The formula for the natural frequency using this method is given by:
f = (0.56 / 2π) * √((E * I) / (m * L^3))
Where:
f is the natural frequency
E is the Young's modulus of the beam material
I is the second moment of area of the section
m is the mass distribution per unit length
L is the span of the beam
In this case, the span of the cantilever beam is given as 8 m, the mass distribution is 200 kg/m, the second moment of area is 4×10^(-4) m^4, and the Young's modulus is 200 GPa.
Substituting these values into the formula, we have:
f = (0.56 / 2π) * √((200 GPa * 4×10^(-4) m^4) / (200 kg/m * (8 m)^3))
Simplifying the equation, we find:
f ≈ 0.056 Hz
Therefore, the natural frequency of transverse vibrations of the cantilever beam, estimated using Dunkerley's empirical method, is approximately 0.056 Hz.
Learn more about Dunkerley's empirical method here:
https://brainly.com/question/32619604
#SPJ11
At each point on the surface of the cube shown in the figure the electric field is parallel to the \( z \) axis. The length of each edge of the cube is \( 3.7 \mathrm{~m} \). On the top face of the cu
Given,Length of each edge of the cube = 3.7 m
Electric field is parallel to the z-axis.If the electric field is only parallel to the z-axis, it means the electric field is directed along the z-axis and has no components along the x-axis and the y-axis.
Therefore, the component of the electric field, E that passes through the surface of the cube is given by
E = Ecosθ
Where θ = 0° since the electric field is parallel to the z-axis.
On the top face of the cube, the direction of the normal vector of the surface is along the negative z-axis since the electric field is passing from top to bottom.
Therefore, the angle between the electric field and the surface on the top face isθ = 180°, and the component of the electric field that passes through the top face isE = Ecos180° = −E
The magnitude of the electric field E is given by the relationE = V/d
where V is the voltage between the top and bottom faces of the cube and d is the distance between the top and bottom faces of the cube.
Substituting the given values,V = 56 Vd
= length of the cube
= 3.7 m
Therefore,E = 56/3.7 = 15.14 V/m
Thus, the magnitude of the electric field is 15.14 V/m.
The negative sign in the answer indicates that the electric field is directed along the negative z-axis.
To know more about edge visit:
https://brainly.com/question/1391344
#SPJ11
A vector
B
has components B
x
=−5 and B
y
=−4 Q1.1 1 Point Determine the magnitude of
B
. 4.60 7 6.4 3 Q1.2 2 Points Determine the angle that
B
makes and state from which axis you are measuring this angle. θ=tan
−1
∣
∣
∣B
y
∣
B
v
∣
∣
=51.3
∘
c.w from +x θ=tan
−1
∣
∣
B
y
B
y
∣
∣
=51.3
∘
c.w from −y θ=tan
−1
∣
∣
B
x
B
y
∣
∣
=38.7
∘
cw from −y θ=tan
−1
∣
∣
∣B
y
B
y
∣
∣
=38.7
∘
c.c.w from −y Q2 2 Points You are given 2 vectors
A
=
i
+
j
and
B
=
i
−
j
. What is the angle in degrees between
A
and
B
? 180 45 90 360
(1.1) option 4, The magnitude of vector [tex]\overrightarrow B[/tex] is 3 and (1.2) option 4, the angle it makes with the negative y-axis is [tex]38.7^0[/tex], measured counter-clockwise. Q2. option 3, The angle between vectors [tex]\overrightarrow A[/tex] and [tex]\overrightarrow B[/tex] is [tex]90^0[/tex].
Q1.1, For determining the magnitude of vector [tex]\overrightarrow B[/tex], use the formula:
[tex]magnitude = \sqrt(B_x^2 + B_y^2)[/tex]
Substituting the given values:
[tex]\sqrt((-5)^2 + (-4)^2) = \sqrt(25 + 16) = \sqrt(41) \approx 6.4[/tex]
Therefore, the magnitude of vector [tex]\overrightarrow B[/tex] is approximately 6.4.
Q1.2, For finding the angle that vector [tex]\overrightarrow B[/tex] makes, use the formula:
[tex]\theta = tan^{(-1)}(|B_x|/|B_y|)[/tex]
Substituting the given values:
[tex]\theta = tan^{(-1)}(|-5|/|-4|) \approx 38.7^0[/tex]
The angle is measured counter-clockwise from the negative y-axis.
Q2, for determining the angle between vectors A and B, use the dot product formula:
A · B = |A| |B| cos(θ).
Since A and B have magnitudes of 1, the formula simplifies to cos(θ) = A · B.
Calculating the dot product of A and B:
A · B = (1)(1) + (1)(-1) = 1 - 1 = 0
Therefore, cos(θ) = 0, which implies that the angle between A and B is [tex]90^0[/tex].
Learn more about magnitude here:
https://brainly.com/question/31022175
#SPJ11
The complete question is:
A vector [tex]\overrightarrow B[/tex] has components [tex]B_x =-5[/tex] and [tex]B_y =-4[/tex]
Q1.1) Determine the magnitude of [tex]\overrightarrow B[/tex].
1. 4.60
2. 7
3. 6.4
4. 3
Q1.2 ) Determine the angle that [tex]\overrightarrow B[/tex] makes and state from which axis you are measuring this angle.
1. θ=[tex]tan^{-1}[/tex]∣B_x∣/∣B_y∣=[tex]51.3^0[/tex] c.w from +x
2. θ=[tex]tan^{-1}[/tex]∣B_x∣/∣B_y∣=[tex]51.3^0[/tex] c.w from -y
3. θ=[tex]tan^{-1}[/tex]∣B_x∣/∣B_y∣=[tex]38.7^0[/tex] c.w from -y
4. θ=[tex]tan^{-1}[/tex]∣B_x∣/∣B_y∣=[tex]38.7^0[/tex] c.c.w from -y
Q.2) You are given 2 vectors [tex]\overrightarrow A= \overrightarrow i+ \overrightarrow j[/tex]and [tex]\overrightarrow B = \overrightarrow i-\overrightarrow j[/tex]. What is the angle in degrees between [tex]\overrightarrow A[/tex] and [tex]\overrightarrow B[/tex]?
1. 180
2. 45
3. 90
4. 360
A ball is projected straight up with an initial velocity of 30 m/s. After 3 seconds its velocity will be 0 m/s. 20 m/s. −20 m/s. 9.4 m/s. −9.4 m/s.
A ball projected upwards with an initial velocity of 30 m/s will not have a velocity of 0, 20, -20, 9.4, or -9.4 m/s after 3 seconds.
To solve this problem, we can use the kinematic equation for velocity:
v = u + at
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
In this case, the ball is projected straight up, so the acceleration is due to gravity and is equal to -9.8 m/s² (assuming no air resistance). The initial velocity (u) is 30 m/s, and we want to find the final velocity (v) after 3 seconds (t = 3 s).
Using the equation, we have:
v = u + at
v = 30 m/s + (-9.8 m/s²)(3 s)
v = 30 m/s - 29.4 m/s
v = 0.6 m/s
Therefore, the velocity of the ball after 3 seconds is approximately 0.6 m/s. None of the given options match this result, so none of the provided velocities are correct in this case.
To know more about velocity, click here:
brainly.com/question/30559316
#SPJ11
When electromagnetic radiation of wavelength 679 nm is incident on a metal of unknown composition, the result is the ejection of electrons with kinetic energies as high as 0.65eV. What must be the binding energy (in eV ) of the metal?
The binding energy of the metal is approximately 6.425 eV.
To determine the binding energy of the metal, we need to use the relationship between the energy of incident photons, the work function of the metal, and the kinetic energy of the ejected electrons.
The energy of a photon can be calculated using the equation:
E = hc/λ
Where:
E is the energy of the photon,
h is the Planck's constant (approximately [tex]6.626 * 10^{-34}[/tex] J·s),
c is the speed of light (approximately [tex]3.00 * 10^8[/tex] m/s), and
λ is the wavelength of the electromagnetic radiation.
The work function of the metal represents the minimum amount of energy required to remove an electron from the metal's surface.
The binding energy (BE) of the metal is related to the work function (Φ) and the kinetic energy (KE) of the ejected electrons:
BE = KE + Φ
Given:
Wavelength (λ) = 679 nm =[tex]679 * 10^{-9}[/tex] m
Kinetic Energy (KE) = 0.65 eV
First, let's calculate the energy of the incident photons using the wavelength:
E = hc/λ
Substituting the values:
[tex]E = \frac {(6.626 * 10^{-34} * 3.00 * 10^8)}{(679 * 10^{-9})}[/tex]
Calculating the value of E:
[tex]E \approx 9.24 * 10^{-19} J[/tex]
To convert this energy value to electronvolts (eV), we can divide it by the elementary charge (approximately 1.6 × 10^(-19) C), which is the charge of one electron:
E (in eV) = E / (1.6 × 10^(-19) C)
Substituting the value of E:
E (in eV) ≈ (9.24 × 10^(-19) J) / (1.6 × 10^(-19) C)
E (in eV) ≈ 5.775 eV
Now we can calculate the binding energy (BE) of the metal using the kinetic energy (KE) and the energy of the incident photons (E):
BE = KE + E (in eV)
Substituting the values:
BE = 0.65 eV + 5.775 eV
Calculating the binding energy:
BE ≈ 6.425 eV.
To know more about binding energy
brainly.com/question/31748572
#SPJ11
Some insects can walk below a thin rod (such as a twig) by hanging from it. Suppose that such an insect has mass m and hangs from a horizontal rod as shown in the figure, with angle θ=35∘
. Its six legs are all under the same tension, and the leg sections nearest the body are horizontal. (a) What is the ratio of the tension in each tibia (forepart of a leg) to the insect's weight? (b) If the insect straightens out its legs somewhat, does the tension in each tibia increase, decrease, or stay the same?
The ratio of the tension in each tibia to the insect's weight is (1/2) * cos(35°). (b) When the insect straightens its legs somewhat, the tension in each tibia decreases.
(a) To find the ratio of the tension in each tibia to the insect's weight, we need to consider the forces acting on the hanging insect. In this case, the weight of the insect is acting vertically downward, while the tension in each tibia is acting along the legs. By analyzing the equilibrium of forces, we can determine the ratio.
To elaborate, we can consider the forces involved. The weight of the insect can be represented by the force acting vertically downward, which is equal to the mass of the insect multiplied by the acceleration due to gravity (m*g). Since all six legs are under the same tension, the total tension force can be divided equally among the six legs. Therefore, the tension in each tibia is equal to one-sixth of the total tension force.
By dividing the tension in each tibia by the weight of the insect, we can calculate the desired ratio. This ratio will provide insights into the relative strength of the insect's legs in supporting its weight while hanging from the rod.
(b) If the insect straightens out its legs somewhat, the tension in each tibia may change. By extending the legs, the angles between the legs and the rod may be altered. This can affect the vertical and horizontal components of the forces acting on the insect's legs. Depending on the specific changes in angles, the tension in each tibia can either increase, decrease, or remain the same. To determine the change in tension, a detailed analysis of the forces and angles involved in the new leg configuration is required.
To learn more about tension, click here: https://brainly.com/question/32990542
#SPJ11
With the aid of fully labeled diagram (s), explain how P-N- P transistor is biased to operate in the active region. Include in your explanation how holes and electrons behave
When a P-N-P transistor is biased in the active region, electrons from the N-type emitter flow into the P-type base, recombine with the holes, and a small fraction of the electrons diffuse through the base into the N-type collector. This flow of electrons from the emitter to the collector constitutes the collector current, which is controlled by the base current.
To understand how a P-N-P transistor is biased to operate in the active region, let's first briefly discuss the behavior of holes and electrons in a semiconductor material.
In a semiconductor material such as silicon, there are two types of charge carriers: electrons and holes. Electrons carry negative charge and are the majority carriers in N-type semiconductors, while holes carry positive charge and are the majority carriers in P-type semiconductors.
Now, let's delve into the operation of a P-N-P transistor biased in the active region. The P-N-P transistor consists of three layers: a P-type layer sandwiched between two N-type layers. The middle P-type layer is called the base, while the N-type layers on either side are called the emitter and collector.
To operate the P-N-P transistor in the active region, we need to bias the transistor properly. This means applying appropriate voltages to the emitter-base and collector-base junctions. Let's assume the base-emitter junction is forward-biased, which means the emitter is at a higher potential than the base.
When a forward bias is applied to the base-emitter junction, electrons from the N-type emitter region begin to flow towards the P-type base region. These electrons recombine with the holes present in the base region. The base is very thin compared to the other regions, allowing for efficient recombination.
However, due to the thinness of the base region, only a small number of electrons actually recombine with the holes. The remaining majority of electrons diffuse through the base and enter the collector region, which is reverse-biased. The reverse bias on the collector-base junction prevents current flow from the collector to the base.
As the electrons diffuse from the emitter to the collector, they form the main current flow through the transistor. This current is called the collector current (Ic). The amount of collector current is controlled by the amount of current flowing into the base-emitter junction, which is the base current (Ib).
The base current is typically much smaller than the collector current, and the transistor is designed to amplify this small base current into a larger collector current. This property of current amplification is what makes transistors useful in electronic circuits.
Learn more about P-N-P transistor from the link given below.
https://brainly.com/question/32902825
#SPJ4
A gas contained within a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2. where P1 = 10 bar, V1 = 0.1m³, U1 - 400 kJ and P2 = 1 bar, V2 = 1.0 m² U2 - 200 kPa: Process A: Process from 1 to 2 during which the pressure-volume relation is PV - constant. Process B: Constant-volume process from state 1 to a pressure of 1 bar, followed by a linear pressure-volume process to state 2. Kinetic and potential energy effects can be ignored. For each of the processes A and B. (a) evaluate the work, in kJ, and (b) evaluate the heat transfer, in kJ.
The heat transfer during process B is -200.49 kJ.
The pressure-volume states of the gas in a piston-cylinder assembly are given as follows:
State 1:
[tex]P_1$ = 10 bar,$V_1$ = 0.1 m^3,$U_1$ = 400 kJ[/tex]
State 2:
[tex]P_2$ = 1 bar,$V_2$ = 1.0 m³,$U_2$ = 200 kJ[/tex]
Process A:
The pressure-volume relation is given by PV - constant, which implies that [tex]$P_1V_1 = P_2V_2$[/tex]. By substituting the given values, we find [tex]$V_2 = 1 m^3$[/tex] .The work done during process A can be calculated as follows:
[tex]$$W = \int\limits^1_2 PdV =\int\limits^1_2 \frac{constant}{V} dV = constant. ln(\frac{V_2}{V_1})$$[/tex]
Hence,
[tex]W = 10 ln(\frac{1} {0.1} ) = 23.03 kJ[/tex]
Since the process is not specified to be adiabatic, heat transfer occurs. According to the first law of thermodynamics, we have:
[tex]$$Q = \Delta U + W$$[/tex]
Substituting the values:
[tex]$$Q = U_2 - U_1 + W = 200 - 400 + 23.03 = -176.97 kJ$$[/tex]
Therefore, the heat transfer during process A is -176.97 kJ.
Process B:
The gas undergoes a constant-volume process from state 1 to a pressure of 1 bar, followed by a linear pressure-volume process to reach state 2. The volume during the constant-volume process is V_1 = 0.1 m³. We can calculate the pressure during this process as follows:
[tex]$$P_2 = P_1(\frac{V_1}{V_2})^{\gamma} = 10(\frac{0.1}{1})^{1.4} = 3.71 bar$$[/tex]
The work done during the constant-volume process is zero since the volume remains constant. For the linear pressure-volume process, the relation is given by PV = constant. Using the given states, we can find the value of the constant:
[tex]$$P_1V_1 = P_2V_2 \Rightarrow 10 \times 0.1 = 3.71 \times V_2 \Rightarrow V_2 = 0.27 m^3$$[/tex]
The work done during the linear pressure-volume process is calculated as the area under the process curve. It can be determined as:
[tex]W = \frac{(P_1 - P_2)(V_1 - V_2)}{2} = \frac{(10 - 3.71)(0.1 - 0.27)} {2} = -0.49 kJ$$[/tex]
Again, since the process is not specified to be adiabatic, there is heat transfer. Applying the first law of thermodynamics, we have:
[tex]$$Q = \Delta U + W = U_2 - U_1 + W = 200 - 400 + (-0.49) = -200.49 kJ$$[/tex]
Therefore, the heat transfer during process B is -200.49 kJ.
Learn more about heat transfer
https://brainly.com/question/13433948
#SPJ11
Three point charges lie along a straight line as shown in the figure below, where q
1
=6.12μC,q
2
=1.51μC, and q
3
=−1.92μC. The separation distances are d
1
=3.00 cm and d
2
=2.00 cm. Calculate the magnitude and direction of the net electric force on each of the charges. (a) q
1
magnitude direction (b) q
2
magnitude direction (c) q
3
magnitude direction
Direction of the net electric force on q1 will be opposite and Direction of the net electric force on q2 will be attractive.
To calculate the net electric force on each of the charges, we can use Coulomb's law, which states that the magnitude of the electric force between two point charges is given by:
F = k × |q1 × q2| / r^2
Where:
F is the magnitude of the electric force.
k is the electrostatic constant, approximately equal to 8.99 × 10^9 Nm^2/C^2.
q1 and q2 are the magnitudes of the charges.
r is the separation distance between the charges.
(a) Net electric force on q1:
The electric force on q1 due to q2 can be calculated using Coulomb's law:
F12 = k × |q1 × q2| / d1^2
Substituting the values:
F12 = (8.99 × 10^9 Nm^2/C^2) × |(6.12 × 10^-6 C) × (1.51 × 10^-6 C)| / (0.03 m)^2
Calculating this, we find:
F12 = 1.830 N
The direction of the force will be attractive since q1 and q2 have opposite charges.
(b) Net electric force on q2:
To find the net electric force on q2, we need to consider both q1 and q3.
Force due to q1:
F21 = k × |q1 × q2| / d1^2
F21 = (8.99 × 10^9 Nm^2/C^2) * |(6.12 × 10^-6 C) * (1.51 × 10^-6 C)| / (0.03 m)^2
Force due to q3:
F23 = k × |q2 × q3| / d2^2
F23 = (8.99 × 10^9 Nm^2/C^2) × |(1.51 × 10^-6 C) × (1.92 × 10^-6 C)| / (0.02 m)^2
The net force on q2 is the vector sum of F21 and F23, which can be calculated using vector addition. The direction will depend on the relative magnitudes and directions of these forces.
(c) Net electric force on q3:
The force on q3 due to q2 can be calculated using Coulomb's law:
F32 = k × |q2 × q3| / d2^2
F32 = (8.99 × 10^9 Nm^2/C^2) × |(1.51 × 10^-6 C) × (1.92 × 10^-6 C)| / (0.02 m)^2
The direction of the force will be attractive since q2 and q3 have opposite charges.
Learn more about electric force -
brainly.com/question/30236242
#SPJ11
