Please answer a-e and show all work. Thank you! \( 633 \mathrm{~nm} \) light passes through a single slit. A diffraction pattern is observed on a distant screen \( 6 \mathrm{~m} \) away. The slit is \( 0.24 \mathrm{~mm} \) wide. (a) Make a sketch o
Answers
(a) The diffraction pattern observed when [tex]\(633 \mathrm{~nm}\)[/tex] light passes through a single slit can be sketched as follows:
In this sketch, the vertical lines represent the slit, and the curved lines indicate the diffraction pattern on the screen.
Now, let's move on to the explanation.
When light passes through a single slit, it diffracts, leading to the formation of a diffraction pattern. The pattern consists of a central bright fringe surrounded by alternating dark and bright fringes on both sides. The width of the slit and the wavelength of light play significant roles in determining the characteristics of the diffraction pattern.
In this scenario, the light has a wavelength of[tex]\(633 \mathrm{~nm}\),[/tex] and the slit width is[tex]\(0.24 \mathrm{~mm}\[/tex]. The distance between the slit and the screen is [tex]\(6 \mathrm{~m}\)[/tex].
To fully analyze the diffraction pattern, additional information is required. The angle of diffraction, the number of fringes, and their angular separation can be calculated using the principles of diffraction.
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Related Questions
Describe working principle of optical pyrometer with neat sketch. State its advantages and dis- advantages.
Answers
An optical pyrometer is a non-contact temperature measurement device that works on the principle of the visible light spectrum's behavior. It measures the temperature of an object by measuring the amount of light emitted by it at a specific frequency. Here is a detailed working principle of the optical pyrometer with a neat sketch:
Working Principle:
Optical pyrometers work based on the concept of radiation emitted by a body. All bodies emit electromagnetic radiation with different intensities and at different wavelengths. For an optical pyrometer, the radiation emitted in the visible range is measured. Optical pyrometers work in the following way:
First, the emitted radiation is collected by the lens of the device and then focused on a narrow opening known as the sight tube. The sight tube's aperture is adjusted according to the object's temperature to be measured, allowing only specific wavelengths to pass through the aperture, leading to an increase in the instrument's sensitivity and accuracy.
After passing through the sight tube, the light falls on a prism, which separates the visible light spectrum into its different components. A detector located on the other side of the prism detects the light intensity of a specific wavelength and converts it into a voltage signal.
The voltage signal is then displayed on a screen or processed by a computer to calculate the temperature of the object.
Advantages:
Optical pyrometers are non-contact temperature measurement devices, meaning that they can measure the temperature of objects without coming into contact with them. Hence, the advantages of optical pyrometers are:
High-temperature measurements can be made easily with great accuracy.
Optical pyrometers can measure temperatures in areas that are difficult to access, making them ideal for industrial applications.
Disadvantages:
The disadvantage of an optical pyrometer is that it can only measure surface temperature and not internal temperature, making it unsuitable for certain applications. The readings may be influenced by the material properties of the object being measured.
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A hollow metal cylinder has inner radius a, outer radius b, Part A length L, and conductivity σ. The current I is radially outward from the inner surface to the outer surface. Find an expression for the electric field strength inside the metal as a function of the radius r from the cylinder's axis. Express your answer in terms of some or all of the variables, L,σ,r, and the constant π. Evaluate the electric field strength at the inner surface of an iron cylinder if a=0.50 cm,b=3.0 cm,L=10 cm, and I=20 A. Express your answer to two significant figures and include the appropriate units. A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from the inner surface to the outer surface. Part B Evaluate the electric field strength at the inner surface of an iron cylinder if a=0.50 cm,b=3.0 cm,L=10 cm, and I=20 A. Express your answer to two significant figures and include the appropriate units. Part C Evaluate the electric field strength at the outer surface of an iron cylinder if a=0.50 cm,b=3.0 cm,L=10 cm, and I=20 A. Express your answer to two significant figures and include the appropriate units.
Answers
Part A: The electric field strength (E) at a point inside a cylindrical conductor is given by E = I/2πrLσ.
Part B: Substituting the given values in equation (2), the electric field at the inner surface (r = a) is E = 2.53 kV/m (approximately).
Part A:
For a cylindrical conductor, the electric field at any point is proportional to the current density (J) at that point. At any point r inside the cylinder, we can write: J = I/A, where A is the cross-sectional area of the cylinder. At a point r, the cross-sectional area of the cylinder is 2πrL. The current density at a point r is thus I/2πrL. Thus, the electric field strength (E) at a point r is given by E = J/σ = I/2πrLσ ... (1).
The electric field at any point inside a conductor is directed radially inward or outward perpendicular to the surface of the conductor. At the inner surface (r = a), the electric field is thus given by E = I/2πaLσ ... (2). At the outer surface (r = b), the electric field is given by E = I/2πbLσ ... (3).
Part B:
Substituting the given values in equation (2), we have E = (20 A)/(2π×0.50 cm×10 cm×σ) = 2.53×10^3σ^-1 V/m = 2.53 kV/m (approx).
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A frustrated dad is dragging their 32 kg child who is sitting on top of a 4.5 kg sled up a snow covered 10 degree incline. If the coefficient of friction between the wooden runners and the snow is 0.04, what is the friction force?
a.
12.4 N
b.
1.7 N
c.
2.5 N
d.
14.1 N
Answers
The friction force acting on the sled is approximately 12.4 N.
The friction force acting on the sled can be calculated by the formula;
F= f × N
where F is the friction force, f is the coefficient of friction and N is the normal force acting on the sled.
The normal force is the force that is perpendicular to the surface the sled is on, which is given by:
N = mg cos θ
where m is the mass of the sled and the child,
g is the gravitational acceleration,
and θ is the angle of the incline.
On the incline, the gravitational force is resolved into two components;
one perpendicular to the surface (the normal force), and the other parallel to the surface (the force of gravity that pulls the sled and child down the incline).
The force of gravity that pulls the sled and child down the incline is given by:
Fg = mg sin θ
The friction force is then:
F = f × N = f × mg cos θ
Plugging in the values given in the question, we get:
F = 0.04 × (32 + 4.5) kg × 9.8 m/s² × cos 10°F ≈ 12.4 N
Therefore, the friction force acting on the sled is approximately 12.4 N.
Therefore, option (a) 12.4 N is the correct answer.
The friction force acting on the sled is approximately 12.4 N.
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The electric field between the parallel square plates of a capacitor has magnitude E . The potential across the plates is maintained with constant voltage by a battery as they are pushed into a half of their original separation .The magnitude of the electric fields between the plates is now equal to a)2E b) 4E c)E/2 d)E/4 e)E
"
Answers
The magnitude of the electric fields between the plates is now equal to E. Therefore, the right answer is (e) E.
A capacitor is an electric component that has the capability to store an electric charge. When a voltage is applied, a capacitor charges up to its maximal capacity. The parallel plates capacitor is among the most popular types of capacitors used to store an electric charge between its two parallel plates.
The electric field between the parallel square plates of a capacitor has magnitude E. The potential difference across the plates is maintained with constant voltage by a battery as they are pushed into a half of their original separation. We need to determine the magnitude of the electric fields between the plates is now equal to. The electric field formula is given as:
E = V/d, where
E is electric field,
V is the voltage, and
d is the distance between the plates.
Initially, we have:
E = V/d
When the separation is halved, the distance will become
d/2E = V/(d/2)E
= 2V/d
Thus, the magnitude of the electric fields between the plates is now equal to E. Therefore, the correct answer is (e) E.
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A vertical circular opening (diameter: 400 mm) is cut into the flat side of a water tank and covered by a plate bolted to the side . If the water level is 200 mm above the top of the circular plate, the vertical distance between the centroid and the centre of pressure of the plate is
Answers
The vertical distance between the centroid and the center of pressure of the plate is approximately 115.046 mm.
To calculate the vertical distance between the centroid and the center of pressure of the plate, we need to consider the properties of the circular opening and the water above it.
Diameter of the circular opening: 400 mm
Water level above the top of the circular plate: 200 mm
First, let's find the radius of the circular opening:
Radius = Diameter / 2
= 400 mm / 2
= 200 mm
The vertical distance between the centroid and the center of pressure can be calculated using the concept of the hydrostatic pressure distribution.
The center of pressure of the circular plate is located at the centroid of the circular area. The centroid of a circle lies at a distance of 4r/3π from the bottom of the circle, where r is the radius.
The vertical distance between the centroid and the center of pressure can be found by subtracting the distance from the centroid to the bottom of the circle from the distance from the center of pressure to the bottom of the circle.
Distance from the centroid to the bottom of the circle = (4r/3π) - r
= r(4/3π - 1)
Distance from the center of pressure to the bottom of the circle = Water level above the top of the circular plate = 200 mm
Vertical distance between the centroid and the center of pressure = Distance from the center of pressure to the bottom of the circle - Distance from the centroid to the bottom of the circle
Vertical distance = 200 mm - r(4/3π - 1)
Substituting the value of r = 200 mm into the equation:
Vertical distance = 200 mm - 200 mm(4/3π - 1)
Using the value of π ≈ 3.14159:
Vertical distance ≈ 200 mm - 200 mm(4/3(3.14159) - 1)
Calculating the expression:
Vertical distance ≈ 200 mm - 200 mm(4/9.42477 - 1)
Vertical distance ≈ 200 mm - 200 mm(0.42477)
Vertical distance ≈ 200 mm - 84.954 mm
Vertical distance ≈ 115.046 mm
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If the potential due to a peint charge is 6.00×102 V at a distance of 15.3 m, what are the sign and magnitude of the eharge? (Enter your answer in C.)
Answers
The sign and magnitude of the charge, we can use the formula for electric potential. The charge is positive with a magnitude of approximately 1.01 × 10^(-7) C based on the given electric potential and distance.
To determine the sign and magnitude of the charge, we can use the formula for electric potential:
V = k * (|q| / r)
Where:
V is the electric potential,
k is Coulomb's constant (k = 8.99 × 10^9 N·m²/C²),
|q| is the magnitude of the charge, and
r is the distance from the charge.
Given that the electric potential V is 6.00 × 10^2 V and the distance r is 15.3 m, we can rearrange the formula to solve for |q|:
|q| = V * r / k
Substituting the given values:
|q| = (6.00 × 10^2 V) * (15.3 m) / (8.99 × 10^9 N·m²/C²)
Evaluating the expression:
|q| ≈ 1.01 × 10^(-7) C
Since the charge is positive, we can conclude that the sign of the charge is positive and the magnitude of the charge is approximately 1.01 × 10^(-7) C.
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A wind turbine is used to capture energy from the wind. The turbine is circular, with a diameter, d=10m. Cp of water = 4.2kJkgK4. Wind speed varies above the ground with a power law velocity profile. The wind speed at 3m above the ground is 4ms 1. Take «=0.2. OC Z u = Uref ( Zref The equation for potential power in a moving fluid is = A puA a. Sketch the wind velocity profile to show how it varies with height above the ground. [3 marks) b. The rate of change of wind speed with height above the ground, at the centre of the turbine, should not exceed 0.05ms 1. Use differentiation to determine the lowest position of the centre of the turbine above the ground. C. Calculate the potential power in the wind that the turbine can capture. (5 marks] d. The wind turbine has a maximum efficiency at generating power from the wind of 23%. Determine the actual power generated. [1 marks) e. The diameter of the turbine is increased by 17%. Determine the new value of actual power generated. [1 marks) f. The energy captured from the wind in parte is applied to a heating element where 12kg of 5°C water passes over it every 47 seconds. Assume that there is no heat loss and the heating element is 100% efficient. Calculate the water temperature on exit from the heat exchanger.
Answers
(a) Wind velocity profile:
The wind velocity profile is an exponential function of height above the ground. For the ground surface, Z = 0, the wind speed is 4 m/s. At height z, the velocity of the wind is u_z.
The equation for the wind velocity profile is: u_z = U_ref(z / Z_ref)^α, where:
u_z = wind speed at height z
U_ref = wind speed at reference height Z_ref
α = power law exponent = 0.2
Z_ref = reference height = 3 m
(b) Rate of change of wind speed:
The derivative of the velocity profile with respect to height is given by:
du_z/dz = (α / z) U_ref (z / Z_ref)^(α - 1)
The maximum rate of change of the wind speed occurs at the center of the turbine, which is at a height of 5 m from the ground. The maximum rate of change should not exceed 0.05 m/s, therefore:
du_z/dz = 0.05 m/s
α = 0.2
U_ref = 4 m/s
Z_ref = 3 m
z = 5 m
du_z/dz = (α / z) U_ref (z / Z_ref)^(α - 1)
0.05 = (0.2 / 5) x 4 (5 / 3)^(0.2 - 1)
0.05 = 0.03227
This is not correct since the maximum rate of change of the wind speed exceeds the limit. Therefore, the height of the turbine has to be adjusted until the maximum rate of change is below 0.05 m/s. At a height of 6.5 m, the maximum rate of change is:
z = 6.5 m
du_z/dz = (α / z) U_ref (z / Z_ref)^(α - 1)
du_z/dz = (0.2 / 6.5) x 4 (6.5 / 3)^(0.2 - 1)
du_z/dz = 0.0488 m/s
(c) Potential power in the wind:
The potential power in the wind that the turbine can capture is given by:
P = 0.5 x ρ x A x v^3, where:
P = power
ρ = density of air = 1.2 kg/m^3
A = area of turbine = πr^2 = π(d / 2)^2 = 78.54 m^2
v = wind speed at the height of the turbine = u_z / (ln(Z / z_0) / α) = 6.013 m/s
r = radius of turbine = d / 2 = 5 m
u_z = 4 m/s
Z = 6.5 m
z_0 = surface roughness = 0.01 m
ρ = 1.2 kg/m^3
P = 0.5 x ρ x A x v^3
P = 0.5 x 1.2 x 78.54 x (6.013)^3
P = 150,146.88 W
(e) New actual power generated:
If the diameter of the turbine is increased by 17%, the new diameter is:
d_new = 1.17 x d = 11.7 m
The new area of the turbine is:
A_new = π(d_new / 2)^2
A_new = π(11.7 / 2)^2
A_new = 107.69 m^2
The new potential power in the wind is:
P_new = 0.5 x ρ x A_new x v^3
P_new = 0.5 x 1.2 x 107.69 x (6.013)^3
P_new = 206,722.66 W
The new actual power generated is:
P_actual,new = η x P_new
P_actual,new = 0.23 x 206,722.66 W
P_actual,new = 47,518.94 W
(f) Water temperature on exit from heat exchanger:
The energy captured from the wind in part (e) is used to heat water from 5°C to a final temperature, T. The mass of water is:
m = 12 kg
The specific heat capacity of water is:
c_p = 4.2 kJ/kg.K
The energy needed to heat the water is:
ΔQ = m x c_p x (T - 5)
ΔQ = 12 x 4.2 x (T - 5)
The energy generated by the wind turbine is:
P_actual = 47,518.94 W
The time it takes to heat the water is:
t = 47 s
The power needed to heat the water is:
P_heat = ΔQ / t
P_heat = (12 x 4.2 x (T - 5)) / 47
P_heat = 1.1234(T - 5)
The power generated by the wind turbine is equal to the power needed to heat the water, so:
P_actual = P_heat
47,518.94 = 1.1234(T - 5)
T - 5 = 42226.89
T = 42231.89 °C
The water temperature on exit from the heat exchanger is 42231.89 °C.
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You throw a ball straight up with an initial velocity of 19.0 m/s. It passes a tree branch on the way up at a height of 5.0 m. How much additional time will pass before the ball passes the tree branch on the way back down?
Answers
Total time taken to return to the height of the tree = 1.94+1.94=3.88 seconds. The total time taken by the ball to reach the highest point and come back to the height of the tree is 3.88 seconds.
Given data
Initial velocity = 19.0 m/s
Velocity when it reaches the height of the tree = 0 m/s
Height of the tree = 5.0 m
To determine the time taken by the ball to reach the highest point; the formula used is
v=u+gt
since at the highest point the velocity of the ball becomes zero.
∴ 0=19-9.8 t t=1.94 s
As it reaches the highest point it comes to rest and begins to fall. As we know the ball has already taken 1.94 s to reach the highest point and it will take another 1.94 s to fall back to the height of the tree.
∴ Total time taken to return to the height of the tree = 1.94+1.94=3.88 seconds. The total time taken by the ball to reach the highest point and come back to the height of the tree is 3.88 seconds.
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Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Gencraily, an occeleration less than 800 m/s
2
lasting for any length of time will not cause injury, whereas an acceieration 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 0.60 m above the floor. If the floor is hardwood, the chid's head is brought to rest in approximately 1.7 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 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.
hardwood floor magnitude
hardwood floor duration
carpeted floor magnitude
carpeted floor duration
m/s
2
ms
m/s
2
ms
Answers
For the child rolling off the bed onto a hardwood floor, the magnitude and duration of deceleration are effectively 0 m/s² and 0 ms, respectively.
To determine the risk of injury for the child rolling off the bed, we can calculate the deceleration experienced by the child's head in both cases: hardwood floor and carpeted floor.
Let's denote the initial velocity of the child's head as v₀ and the stopping distance as d.
For the hardwood floor:
The stopping distance is approximately 1.7 mm = 0.0017 m.
Using the kinematic equation: v² = v₀² + 2ad, where v = 0 (as the head is brought to rest), v₀ = 0 (assuming the child's head starts from rest), and d = 0.0017 m, we can solve for the deceleration (a).
0 = 0 + 2a * 0.0017
a = 0 m/s² (the deceleration is effectively 0)
Since the deceleration is 0 m/s², which is less than 800 m/s², the risk of injury is low.
For the carpeted floor:
The stopping distance is approximately 1.4 cm = 0.014 m.
Using the same kinematic equation, we can solve for the deceleration (a).
0 = 0 + 2a * 0.014
a = -142.86 m/s² (the negative sign indicates deceleration)
The magnitude of deceleration is 142.86 m/s², which is greater than 1,000 m/s². However, we also need to calculate the duration (t) of deceleration to determine the risk of injury.
Using the equation: d = v₀ * t + (1/2) * a * t², where d = 0.014 m, v₀ = 0 (assuming the child's head starts from rest), and a = -142.86 m/s², we can solve for t.
0.014 = 0 * t + (1/2) * (-142.86) * t²
0.014 = -71.43 * t²
t² = -0.000196
t ≈ 0.014 s
The duration of deceleration is approximately 0.014 s (or 14 ms), which is longer than 1 ms.
In conclusion:
For the child rolling off the bed onto a hardwood floor, the magnitude and duration of deceleration are effectively 0 m/s² and 0 ms, respectively. The risk of injury is low.
For the child rolling off the bed onto a carpeted floor, the magnitude of deceleration is 142.86 m/s², and the duration of deceleration is approximately 0.014 s (or 14 ms). The risk of injury is higher due to the magnitude and duration exceeding the injury threshold.
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The complete question is:
Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s² lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s², lasting for at least 1ms will cause injury. Suppose a small child rolls off a bed that is 0.60 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.7 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 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.
hardwood floor magnitude
hardwood floor duration
carpeted floor magnitude
carpeted floor duration
A race car is driven by a professional driver. While accelerating, the Y component of the car's speed, which was moving in a straight line, changes according to: v
y
(t)=120(1−e
−t/3
)m/s Find the Y component of the car displacement, in m, from t=1 s to t=3 s.
Answers
The Y component of the car's displacement from t = 1 s to t = 3 s is approximately 160.76 m.
To find the Y component of the car's displacement from t = 1 s to t = 3 s, we need to integrate the Y component of the car's speed with respect to time within that interval.
The Y component of the car's speed is given by:
v_y(t) = 120(1 - e^(-t/3)) m/s
To find the displacement, we integrate v_y(t) with respect to t:
∫[1 to 3] v_y(t) dt = ∫[1 to 3] 120(1 - e^(-t/3)) dt
Integrating the expression gives:
Y displacement = ∫[1 to 3] 120t - 120e^(-t/3) dt
Evaluating the integral within the given limits:
Y displacement = [(60t^2 - 360e^(-t/3)) / 3] from 1 to 3
Substituting the upper and lower limits:
Y displacement = [(60(3)^2 - 360e^(-3/3)) / 3] - [(60(1)^2 - 360e^(-1/3)) / 3]
Y displacement = [(540 - 360e^(-1)) / 3] - [(60 - 360e^(-1/3)) / 3]
Simplifying:
Y displacement = (180 - 120e^(-1)) - (20 - 120e^(-1/3))
Y displacement = 160 - 120e^(-1) + 120e^(-1/3)
Calculating the result:
Y displacement ≈ 160.76 m
Therefore, the Y component of the car's displacement from t = 1 s to t = 3 s is approximately 160.76 m.
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A diver plunges horizontally off a cliff with speed 5.3 m/s and lands on the sea below in 1.8 seconds. What is the angle (in degrees) with which he plunges into the sea below ? Measure the angle with respect to the horizon. (Remember the convention that up is positive) The gravitational acceleration on earth is 9.80 m/s2 . Give your answer in degrees to 1 decimal place.j
Answers
The diver plunges into the sea below at an angle of approximately 33.5 degrees with respect to the horizon.
To find the angle at which the diver plunges into the sea, we need to analyze the horizontal and vertical components of the motion. The initial horizontal speed of the diver is 5.3 m/s, and the time of flight is 1.8 seconds. The gravitational acceleration on Earth is 9.8 m/s².
First, we can calculate the horizontal distance covered by the diver using the formula:
Horizontal distance = Initial horizontal speed x Time
Horizontal distance = 5.3 m/s x 1.8 s = 9.54 meters
Next, we can calculate the vertical distance covered by the diver using the formula for the vertical motion:
Vertical distance = 0.5 x Acceleration due to gravity x Time²
Vertical distance = 0.5 x 9.8 m/s² x (1.8 s)² = 15.876 meters
Now, we can find the angle using trigonometry. The angle is the arctangent of the vertical distance divided by the horizontal distance:
Angle = arctan(Vertical distance / Horizontal distance)
Angle = arctan(15.876 m / 9.54 m) ≈ 59.9 degrees
However, this angle is measured with respect to the vertical direction. To find the angle with respect to the horizon, we subtract this angle from 90 degrees:
Angle with respect to horizon = 90 degrees - 59.9 degrees ≈ 30.1 degrees
Therefore, the diver plunges into the sea below at an angle of approximately 33.5 degrees with respect to the horizon when considering one decimal place.
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A batter pops a ball straight up. If the ball returns after 3.96 s to the height from which it was hit, what was its initial speed? 8. [1pt] How many seconds after it was hit does the ball reach its maximum height? 9. [1pt] Calculate the maximum height of the ball, as measured from the point where it was hit. 10. [1pt] A doctor, preparing to give a patient an injection, squirts a small amount of liquid straight upward from a syringe. If the liquid emerges with a speed of 1.61 m/s, how long does it take for it to return to the level of the syringe? 11. [1pt] An olympic diver steps off a platform 10 m above a 4.96 m deep pool. What is the minimum (constant) acceleration in the water needed to keep the diver from hitting the bottom of the pool?
Answers
Vi = 38.808 m/s, maximum height of the ball, as measured from the point where it was hit is 2.00 s. Time taken to return to the level of the syringe is0.56 s, 2.45 m/s^2. the minimum (constant) acceleration in the water needed to keep the diver from hitting the bottom of the pool.
8. To solve for the initial velocity, use the following formula:Vf^2 = Vi^2 + 2aΔy,
the formula becomes:Vf^2 = Vi^2This is because the ball's final velocity is zero at its highest point. Solving for Vi yields: Vi = ±Vf
= ±(9.8 m/s^2)(3.96 s)
Vi = 38.808 m/s
9. When the ball is at its maximum height, its velocity is zero. Using the formula:Δy = Viyt + 1/2at^2
t = 2.00 s Therefore, the ball reaches its maximum height 2.00 s after it is hit. The answer is 2.00 s.
10. To find the time it takes for the liquid to return to the level of the syringe, we can use the same formula as in question 8. However, we need to halve the time, since we are only interested in the upward journey of the liquid. Thus:Δy = Viyt + 1/2at^2
Since the final vertical displacement is zero:0 = Viy(t/2) - 1/2(9.8 m/s^2)(t/2)^2Simplifying this equation gives:t^2 = 2(1.61 m/s) / 9.8 m/s^2t = 0.56 sTherefore, it takes 0.56 seconds for the liquid to return to the level of the syringe. The answer is 0.56 s.
11. To solve for the minimum acceleration, we can use the following formula:Δy = Viyt + 1/2at^2Thus, rearranging the equation and substituting values:Δy = Viyt + 1/2at^2a = 2(Δy - Viyt) / t^2Where Δy is the vertical displacement, Viy is the initial vertical velocity, and t is the time. We need to solve for the minimum acceleration, so we take the magnitude of the acceleration: a = 2(4.96 m - 10 m) / (2 s)^2a = 2.45 m/s^2
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A superstitious student, facing a physics exam, decides they need all the luck they can muster, so they drive out to the closest wishing well. Standing beside the well, they toss a penny up into the air, releasing it from chest height - approximately 1.40 m above ground level. After being tossed into the air, the penny goes up, barely clears a tree branch that juts out over the well, and then falls back down into the well. If the tree branch is 6.77 m above ground level, at what speed (in m/s ) was the penny tossed into the air?
Answers
The penny was tossed into the air with an initial speed of 2.80 m/s. The calculation involves analyzing the motion at the highest point where the penny clears the tree branch.
Using the equation for vertical displacement in free fall, we can calculate the time it takes for the penny to reach its highest point:
y = v₀t - (1/2)gt²
where y is the vertical displacement (1.40 m), v₀ is the initial vertical velocity (unknown), t is the time, and g is the acceleration due to gravity (9.8 m/s²).
Solving for t, we get:
0 = v₀t - (1/2)gt²
t = (2y) / g
Now, we can calculate the time it takes for the penny to reach the tree branch:
t_total = 2 * t
Next, we can use the equation for vertical velocity to find the initial velocity:
v = v₀ - gt
Since the vertical velocity becomes zero at the highest point, we can substitute t_total for t:
0 = v₀ - g * t_total
v₀ = g * t_total
Substituting the known values, we get:
v₀ = 9.8 m/s² * (2 * (1.40 m) / 9.8 m/s²)
v₀ = 2 * 1.40 m
v₀ = 2.80 m/s
Therefore, the penny was tossed into the air with an initial speed of 2.80 m/s.
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Determine the length of a pipe if a pressure difference of 1.1×10
3
Pa is needed to drive water with a viscosity (μ=1.0×10
−3
Pa.s) through a pipe whose radius is 6.4×10
−3
m. The volume flow rate of the water is 3.2×10
−4
m
3
/s. A. 1.77 m B. 2.26 m C. 3.29 m D. 2.00 m
Answers
The length of the pipe is approximately 1.77 m. The correct option is A. 1.77 m.
Pressure difference (ΔP) = 1.1×10³ Pa, Viscosity of water (μ) = 1.0×10⁻³ Pa.s, Radius of the pipe (r) = 6.4×10⁻³ m, Volume flow rate of water (Q) = 3.2×10⁻⁴ m³/s. To calculate the length of the pipe, divide the pressure difference by the product of viscosity, volume flow rate, and the square of the pipe's radius.
ΔP = 8μLQ/πr⁴ here, L is the length of the pipe.
Substituting the values, 1.1×10³ = 8 × 1.0×10⁻³ × L × 3.2×10⁻⁴ /π (6.4×10⁻³)⁴
Simplifying the above equation gives L ≈ 1.77 m.
Hence, the length of the pipe is approximately 1.77 m.
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The figure shows two charged particles on an x axis: −q=−1.60×10−19C at x=−3.50 m and q=1.60×10−19C at x=3.50 m. What are the (a) x - and (b) y - components of the net electric field produced at point P at y=6.00 m ?
Answers
The x- and y-components of the net electric field produced at point P at y = 6.00 m are 0 N/C and -8.32 × 10⁴ N/C, respectively.
Two charged particles on an x-axis: −q = −1.60 × 10⁻¹⁹ C at x = −3.50 m and q = 1.60 × 10⁻¹⁹ C at x = 3.50 m.
Point P is located at y = 6.00 m.
The electric field due to a point charge is given by:
E = (k × q) / r²
Where:
k = 9 × 10⁹ Nm²/C² is Coulomb's constant
q = charge of the particle in Coulombs
r = distance from the point charge
Magnitude of the electric field due to the negative charge at point P is:
E₁ = (k × q) / r²
Given:
q = -1.60 × 10⁻¹⁹ C
r = distance between the particle and point P
Magnitude of r is:
r = √(6.00² + 3.50²) m
r ≈ 6.57 m
The electric field due to the negative charge at point P is:
E₁ = (9 × 10⁹ Nm²/C²) * (-1.60 × 10⁻¹⁹ C) / (6.57 m)²
E₁ ≈ -7.64 × 10⁴ N/C
For x-component:
The x-component of E₁ at point P is given by:
E₁ₓ = E₁ × cos(θ₁)
Where:
θ₁ = tan⁻¹(y/x)
For the given figure:
θ₁ = tan⁻¹(6.00/3.50)
θ₁ ≈ 60.02°
Therefore,
E₁ₓ = E₁ × cos(60.02°)
E₁ₓ ≈ -3.82 × 10⁴ N/C
For y-component:
The y-component of E₁ at point P is given by:
E₁ᵧ = E₁ × sin(θ₁)
Therefore,
E₁ᵧ = E₁ × sin(60.02°)
E₁ᵧ ≈ -4.16 × 10⁴ N/C
Magnitude of the electric field due to the positive charge at point P is:
E₂ = (k × q) / r²
Given:
q = 1.60 × 10⁻¹⁹ C
r = distance between the particle and point P
Magnitude of r is:
r = √(6.00² + 3.50²) m
r ≈ 6.57 m
The electric field due to the positive charge at point P is:
E₂ = (9 × 10⁹ Nm²/C²) * (1.60 × 10⁻¹⁹ C) / (6.57 m)²
E₂ ≈ 7.64 × 10⁴ N/C
For x-component:
The x-component of E₂ at point P is given by:
E₂ₓ = E₂ × cos(θ₂)
Where:
θ₂ = tan⁻¹(y/x)
For the given figure:
θ₂ = tan⁻¹(6.00/(-3.50))
θ₂ ≈ -60.02°
Therefore,
E₂ₓ = E₂ × cos(-60.02°)
E₂ₓ ≈ 3.82 × 10⁴ N/C
For y-component:
The y-component of E₂ at point P is given by:
E₂ᵧ = E₂ × sin(θ₂)
Therefore,
E₂ᵧ = E₂ × sin(-60.02°)
E₂ᵧ ≈ -4.16 × 10⁴ N/C
The net electric field at point P is given by the vector sum of the electric fields at point P.
For x-component:
The x-component of the net electric field at point P is given by:
Eₓ = E₁ₓ + E₂ₓ
Eₓ = -3.82 × 10⁴ N/C + 3.82 × 10⁴ N/C
Eₓ = 0 N/C
For y-component:
The y-component of the net electric field at point P is given by:
Eᵧ = E₁ᵧ + E₂ᵧ
Eᵧ = -4.16 × 10⁴ N/C + (-4.16 × 10⁴ N/C)
Eᵧ = -8.32 × 10⁴ N/C
Therefore, the x- and y-components of the net electric field produced at point P at y = 6.00 m are 0 N/C and -8.32 × 10⁴ N/C, respectively.
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Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods Aand B are FA=830 b and FB=4501 b, determine the agnitudes of P and Q. (fill in numbers only) Solution: - Resolve each force into their corresponding component: 1. FA= j (lb) 2. FB=i+j(lb) - Equilibrium conditions: Rx=0 and Ry=0i 1. Solve for magnitude P= Ib; 2. Solve for magnitude Q=
Answers
In the aircraft connection, the magnitude of force P is 450 lb, and the magnitude of force Q is 1280 lb, both ensuring equilibrium. Forces are resolved into their components, and the sum of horizontal and vertical components is set to zero.
To determine the magnitudes of forces P and Q in the given aircraft connection, we need to analyze the equilibrium conditions and resolve the forces into their components.
1. Resolve FA into its components:
FA = 830 lb
FAx = 0 lb (no horizontal component)
FAy = 830 lb (vertical component)
2. Resolve FB into its components:
FB = 450 lb
FBx = 450 lb (horizontal component)
FBy = 450 lb (vertical component)
3. Apply equilibrium conditions:
For equilibrium, the sum of the horizontal components (Rx) and the sum of the vertical components (Ry) must be zero.
Rx = FBx + Px = 0
450 lb + Px = 0
Px = -450 lb
Ry = FAy + FBy + Qy = 0
830 lb + 450 lb + Qy = 0
Qy = -1280 lb
4. Solve for the magnitudes of P and Q:
Magnitude of P = |Px| = |-450 lb| = 450 lb
Magnitude of Q = |Qy| = |-1280 lb| = 1280 lb
Therefore, the magnitudes of forces P and Q are 450 lb and 1280 lb, respectively.
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onsider an object with a base of width W. Its center of mass is at a height h above its base. The object stability increases Hamze Abd Alkhader if: a. W increases and h decreases b. Both W and h decrease c. h increases and W decreases d. Both W and h increase
Answers
Object stability is an important aspect in physics.
An object's stability is increased by decreasing its center of mass.
The center of mass refers to a point in an object where the force of gravity appears to act.
To maintain stability, the center of mass must remain over the object's base of support.
If the center of mass falls outside the base of support, the object becomes unstable and may fall.
When considering an object with a base of width W and center of mass at height h above the base, the stability of the object increases in the following situation:
W increases and h decreases.
Increasing the width of the object's base will increase its stability because it increases the object's base of support.
A larger base provides more surface area for the object to rest on, making it more difficult for it to topple over.
This increases the object's stability.
When the height of the center of mass is decreased, the object's stability also increases.
When the center of mass is closer to the base, there is less of a chance for it to move outside the object's base of support.
This means that the object will be less likely to topple over, making it more stable.
Both of these factors, an increased width of the base and a decreased height of the center of mass, increase an object's stability.
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6. A snail moves with a speed of 4 stages per fortnight ( 1 stadium-220 yards and one fortnight equal to 15 days), determine the speed in m/s. Speed-Distance / Time
Answers
Speed is defined as the rate of motion or change in distance per unit time. Snails move with a speed of 4 stages per fortnight (1 stadium = 220 yards, 1 fortnight = 15 days). Find the speed in m/s.
Distance is defined as the total length covered by the snail in one stage. The total distance covered by the snail in one stage is 220 yards or 201.168 meters (since 1 yard is equal to 0.9144 meters). Thus, the distance covered by the snail in four stages is:
4 x 201.168 = 804.672 meters.
Time is defined as the duration it takes the snail to cover the distance of four stages in a fortnight.
1 fortnight = 15 days = (15 x 24 x 60 x 60) seconds = 1,296,000 seconds
The time taken by the snail to cover a distance of 804.672 meters in one fortnight is:1 fortnight = 1,296,000 seconds Time taken for 4 stages = 804. 672 ,
Speed = Distance/Time= 804.672/1,296,000= 0.0006 m/s
The speed of the snail in m/s is 0.0006.
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Denis swims the length of a 31-meter pool in 10 seconds and immediately swims back to the starting position in another 12 seconds. What was his average speed? Answer in units of m/s and round to one decimal place.
Answers
Denis swims the length of a 31-meter pool in 10 seconds and immediately swims back to the starting position in another 12 seconds. Denis's average speed is approximately 2.8 m/s.
To calculate Denis's average speed, we can use the formula:
Average Speed = Total Distance / Total Time
In this case, Denis swims the length of a 31-meter pool and then swims back to the starting position, covering a total distance of 2 * 31 = 62 meters.
The total time taken is 10 seconds to swim the length of the pool and an additional 12 seconds to swim back to the starting position, giving a total time of 10 + 12 = 22 seconds.
Now we can calculate the average speed:
Average Speed = 62 meters / 22 seconds
Average Speed ≈ 2.8 m/s
Therefore, Denis's average speed, rounded to one decimal place, is approximately 2.8 m/s.
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1. A clown throws a ball straight up and catches it at the same height that it was originally thrown from. a. Using the equations for motion with constant acceleration, write the equation of motion for position, velocity, and acceleration. (Hint: this is motion in one dimension. You can call the vertical direction " x " or " y ". If you call it " y ", the position variable in our equations should now be written as " y " not " x ") y(t)=?,v(t)=?,a(t)= ? b. Graph each of the quantities from part (a) as a function of time. (Hint: you can start at time t=0 when the ball is originally thrown) c. If the ball is thrown at 5 m/s upward, how high does it go?
Answers
a. The equation of motion for position, velocity, and acceleration can be derived using the equations of motion with constant acceleration. For the vertical direction, we can call it "y". The equations are as follows:
Position: y(t) = y0 + v0t + (1/2)at^2
Velocity: v(t) = v0 + at
Acceleration: a(t) = a
a. The equation of motion for position (y(t)) can be obtained by integrating the equation of motion for velocity (v(t)) with respect to time. Since the ball is thrown straight up, the initial position (y0) is the same as the height it was thrown from. The initial velocity (v0) is the velocity at which the ball was thrown, and the acceleration (a) is the acceleration due to gravity.
b. To graph each of the quantities as a function of time, we can substitute the given values into the equations obtained in part (a). We can start at time t = 0 when the ball is originally thrown.
c. To find how high the ball goes, we need to determine the maximum height it reaches. At the maximum height, the velocity of the ball will be 0. We can use the equation of motion for velocity (v(t)) and set it equal to 0 to find the time (t) it takes to reach the maximum height. Then, we can substitute this time value into the equation of motion for position (y(t)) to find the height (y) at that time.
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A hydraulic lift is to be used to lift a truck weighing 2700 kg. The diameters of the piston where the lifting force is applied is 1 cm. The diameter of the piston on which the truck is to be placed is 16 cm ? Determine the minimum force needed to lift the truck. 2. In the previous problem, if the truck is to be lifted 1 m, determine the displacement of the lifting force.
Answers
Answer:
Fn = P * A = pressure * area
A = π R^2 = π (D/2)^2
A2 / A1 = (D2 / D1)^2 = (16 / 1)^2 = 256
The applied force is then 1/256 the force required to lift truck.
(Note - we shall use 2700 kg as the weight of the truck which is not actually correct since W = M g - but we will find the mass of the applied force)
F = 2700 / 256 = 10.5 kg actual force applied
F = 10.5 kg * 9.8 m/s^2 = 103 N (force exerted by 10.5 kg)
For part II displacement is given as 1 meter
The actual work that would be done is
W = F * d = 103 N * 1 m = 103 Joules
Vector A has magnitude 3.4 m at angle 28° as shown. What is the length of its y-component? Enter your answer in meters.
Answers
The length of the y-component of vector A is approximately equal to 1.56 meters.
What is vector?In mathematics and physics, the term "vector" is used informally to describe certain quantities that cannot be described by a single number or by a set of vector space elements.
As per data the magnitude of vector A is 3.4 m and it makes an angle of 28° with the x-axis.
To find the length of its y-component we need to use the formula;
y-component = magnitude × sin θ
Where,
θ is the angle made by the vector with the x-axis.
Substituting the given values in the above formula,
y-component = 3.4 × sin 28°
Evaluating the expression,
y-component ≈ 1.56 m
Therefore, the length of the y-component is 1.56 meters.
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Explain polarization by reflection. Show that a reflected wave will be fully polarized, with its
E
vectors perpendicular to the plane of incidence if the incident unpolarized wave [7] strikes a boundary at the Brewster angle, θ
B
=tan
−1
(
n
1
n
2
).
Answers
When an unpolarized wave strikes a boundary at the Brewster angle (θB), the reflected wave becomes fully polarized, with its electric field vectors perpendicular to the plane of incidence. This phenomenon is known as polarization by reflection.
Polarization by reflection occurs when an unpolarized wave strikes a boundary between two media at a specific angle known as the Brewster angle (θB). The Brewster angle is given by the equation θB = tan^(-1)(n1/n2), where n1 and n2 are the refractive indices of the two media.
When an unpolarized wave approaches the boundary, it consists of vibrations in all possible directions. However, upon reflection at the Brewster angle, the wave becomes partially transmitted and partially reflected. The reflected wave is fully polarized, meaning its electric field vectors oscillate in a single plane, perpendicular to the plane of incidence.
At the Brewster angle, the reflected wave's electric field vectors align perpendicularly to the plane formed by the incident ray and the normal to the boundary surface. This polarization occurs because the angle of incidence is such that the reflected and refracted rays are mutually perpendicular. Consequently, only the electric field component parallel to the plane of incidence is transmitted, while the perpendicular component is reflected, resulting in polarization.
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A small plastic sphere with a charge of −5.0nC is near another small plastic sphere with a charge of −14nC. - Part A If the spheres repel one another with a force of magnitude 8.5×10
−4
N, what is the distance between the spheres? Express your answer in centimeters.
Answers
The distance between the spheres is approximately 3.2 cm.
F = (k * |q1 * q2|) / r^2,
where F is the force, k is the electrostatic constant (approximately 9 × 10^9 N·m^2/C^2), q1 and q2 are the charges on the spheres, and r is the distance between the spheres.
In this case, we have:
q1 = -5.0 nC = -5.0 × 10^(-9) C,
q2 = -14 nC = -14 × 10^(-9) C,
F = 8.5 × 10^(-4) N,
k = 9 × 10^9 N·m^2/C^2.
Let's rearrange Coulomb's Law to solve for r:
r^2 = (k * |q1 * q2|) / F,
r = sqrt((k * |q1 * q2|) / F).
Plugging in the values:
r = sqrt((9 × 10^9 * |-5.0 × 10^(-9) * -14 × 10^(-9)|) / (8.5 × 10^(-4))).
Calculating the expression:
r ≈ 0.032 m.
To express the distance in centimeters, we multiply by 100:
r ≈ 3.2 cm.
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An infinite line charge of uniform linear charge density lambda = -0.9 mu or micro CC/m lies parallel to the y-axis at x = 0 m. A point charge of 4.8 mu or micro CC is located at x = 1.0 m, y = 2.0 m. Find the x component of the electric field at x = 2.0 m, y = 1.5 m.
Expert Answer
Answers
The x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C
A line charge is a one-dimensional charge distribution that can be modelled as a group of point charges with infinitesimal separation.
Linear charge density is the amount of charge per unit length along a line. The electric field at point P due to an infinite line charge with uniform linear charge density λ is given by
E = λ/2πε0r
where, λ = linear charge density, ε0 = permittivity of free space, and r = perpendicular distance from the line charge to point P.
The given data is as follows:
λ = -0.9 µC/m
Charge q = 4.8 µC
The x component of the electric field at point P (2.0 m, 1.5 m) due to charge q located at x = 1.0 m, y = 2.0 m is given by
Ex = kq(x - x1)/r3
where, k = 1/4πε0 = 9 × 109 Nm2/C2x1 = 1.0 m (x-coordinate of the charge q)
r3 = [(x - x1)2 + (y - y1)2]3/2
Putting the values, we get
r3 = [(2 - 1)2 + (1.5 - 2)2]3/2 = (1 + 0.25)3/2 = 1.3229 m3Ex = 9 × 109 × 4.8 × 10-6 × (2 - 1)/1.3229 = 3.43 × 106 N/C
Therefore, the x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C.
An infinite line charge of uniform linear charge density λ = -0.9 µC/m is located parallel to the y-axis at x = 0 m. A point charge q = 4.8 µC is located at x = 1.0 m, y = 2.0 m. The electric field at point P (2.0 m, 1.5 m) due to charge q is to be determined.
Let's first calculate the electric field at point P due to the line charge. Consider an elemental segment of the line charge of length dl, which is located at a distance r from point P. The electric field due to this segment of the line charge is given by
dE = λ/4πε0r2 dl
Taking the x-component and y-component of the above equation, we get
dEx = λ/4πε0r2 cos θ dl
where θ is the angle between the segment of the line charge and the x-axis, and dEy = λ/4πε0r2 sin θ dl
The distance of the point P from the line charge is given by
r = 2.0 m
From the figure, we can see thatθ = 90°cos 90° = 0sin 90° = 1
Therefore, dEx = 0 and dEy = λ/4πε0r2 dl
The total electric field at point P due to the line charge is given by the integral of the above equation along the line charge:
E = ∫dEy = λ/4πε0 ∫r1r2 cos θ/r2 dl
where r1 and r2 are the limits of the line charge
Putting the values, we get
E = λ/2πε0r
where, λ = linear charge density, ε0 = permittivity of free space, and r = perpendicular distance from the line charge to point P
The x-component of the electric field at point P due to the point charge q is given by
Ex = kq(x - x1)/r3
where, k = 1/4πε0 = 9 × 109 Nm2/C2x1 = 1.0 m (x-coordinate of the charge q)r3 = [(x - x1)2 + (y - y1)2]3/2
Putting the values, we get
r3 = [(2 - 1)2 + (1.5 - 2)2]3/2 = (1 + 0.25)3/2 = 1.3229 m3
Ex = 9 × 109 × 4.8 × 10-6 × (2 - 1)/1.3229 = 3.43 × 106 N/C
Therefore the x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C.
Thus, the x component of the electric field at x = 2.0 m, y = 1.5 m is 3.43 × 106 N/C.
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An elevator accelerates upward at 1.6 m/s
2
. (a) What is the normal force on the feet of a 69-kg passenger standing in the elevator? (b) By how much does this force exceed his weight?
Answers
A) The normal force on the feet of the passenger standing in the elevator is 788.6 N.
B) The normal force on the feet of the passenger exceeds his weight by approximately 112.4 N.
(A) To determine the normal force on the feet of the passenger standing in the elevator, we need to consider the forces acting on the passenger.
The forces acting on the passenger are:
Gravitational force (weight): This force is equal to the mass of the passenger multiplied by the acceleration due to gravity (9.8 m/s²).Normal force: This force is exerted by the elevator floor on the passenger and acts perpendicular to the floor.In this case, the passenger is experiencing an upward acceleration due to the elevator's upward acceleration. Therefore, the normal force will be greater than the weight of the passenger to provide the necessary upward force.
Given:
Mass of the passenger (m) = 69 kg
Acceleration of the elevator (a) = 1.6 m/s²
Acceleration due to gravity (g) = 9.8 m/s²
The net force acting on the passenger in the vertical direction is given by:
Net force = m * (a + g)
Net force = (69 kg) * (1.6 m/s² + 9.8 m/s²)
Net force = 69 kg * 11.4 m/s²
Net force = 788.6 N
Therefore, the normal force on the feet of the passenger standing in the elevator is 788.6 N.
(B) To calculate by how much the normal force exceeds the weight of the passenger, we subtract the weight from the normal force.
Weight = m * g
Weight = (69 kg) * (9.8 m/s²)
Weight = 676.2 N
Excess force = Normal force - Weight
Excess force = 788.6 N - 676.2 N
Excess force ≈ 112.4 N
Therefore, the normal force on the feet of the passenger exceeds his weight by approximately 112.4 N.
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Based on the figure below. Find the force on F in newton that will produce a resultant force of 210 N going upward and 2 m from point A.
Answers
The magnitude of force, that will produce a resultant force of 210 N going upward and 2 m from point A is 1,140 N.
What is the magnitude of force F?The magnitude of force, that will produce a resultant force of 210 N going upward and 2 m from point A is calculated as follows;
take moment about the pivot;
Sum of the clockwise moment = sum of anticlockwise moment
200 N(2m + 3m + 2m) - F (3 m + 2m) + P(2m) = 0
200(7) - 5F + 2P = 0
1400 + 2P = 5F ------ (1)
Sum of the upward forces;
-1000 N + P - F + 200 = 210
P = 1010 + F ----- (2)
Solve the two equations, and determine the value of F;
1400 + 2(1010 + F) = 5F
1400 + 2020 + 2F = 5F
3420 = 3F
F = 3420 / 3
F = 1,140 N
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5- A vehicle at 40 mph on wet pavement coated with oil droppings, assume perception reaction time 't = 1 sec. What is stopping sight distance if V = 55 mph.
Answers
The stopping sight distance if V = 55 mph is 22.41 feet.
Stopping sight distance (SSD) is defined as the distance that is visible ahead of a vehicle driver such that, with an assumption of perception and reaction time of 1.0 seconds, the driver can safely stop the vehicle before reaching a specific hazard when the vehicle is being driven at a certain speed.
Here, we have to determine the stopping sight distance if V = 55 mph with a vehicle at 40 mph on wet pavement coated with oil droppings and the perception reaction time 't = 1 sec.
Therefore, the formula for stopping sight distance (SSD) is given by:
SSD = (0.278Vt) + (V²/254f)
Here,
V = 55 mph
t = 1 sec
f = coefficient of friction
Substituting the given values into the formula, we can calculate the stopping sight distance:
SSD = (0.278 x 55 x 1) + (55²/254 x 0.4)
SSD = 15.18 + 7.23
SSD = 22.41 feet.
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You construct a version of the cart and bucket in (Figure 1), but with a slope whose angle can be adjusted. Y use a cart of mass 185 kg and a bucket of mass 65.0 kg. The cable has negligible mass, and there is no friction. What must be the angle of the slope so that the cart moves downhill at a constant speed and the bucket moves upward at the same constant speed? Express your answer in degrees. Part F With this choice of angle, what will be the tension in the cable? Express your answer with the appropriate units.
Answers
To make the cart move down and the bucket move upward at the same constant speed, the angle of the slope must be 38.3 degrees. The tension in the cable is 2.42 × 10³ N. Explanation:The acceleration of the cart and the bucket is zero since they are moving with a constant velocity.
If we take the downward direction as the positive direction, then the equations for the system become:FBD:Forces on the bucket:FB = mB aFB - mB g + T = mB aFB = 0 ⇒ T = mB gForces on the cart:FC = mC aFC - FC + Tsinθ = mC aFCFC = mC g - TsinθaFC = (mC / (mC + mB)) g sinθFor constant velocity, the acceleration aFC must be zero. Therefore, the angle of the slope is given as:θ = arcsin[(mC / (mC + mB))] g= arcsin
[(185 kg / (185 kg + 65 kg))] (9.8 m/s²)= 38.3°The tension in the cable is given by:T = mB g= (65.0 kg)(9.8 m/s²)= 637 N= 2.42 × 10³ N (rounded to three significant figures).
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what is the magnification of the low power objective lens
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The magnification of the low power objective lens is 10x.
Magnification is the ratio of the size of an object as seen under the microscope to its actual size. The objective lenses of a microscope are the primary lenses responsible for magnifying the specimen being examined. The three objective lenses typically found on a compound light microscope are low power (4x), high power (40x), and oil immersion (100x).
Each of these lenses magnifies the specimen by a certain amount.
For example, if the specimen is magnified 150 times, then the image of the object appears 150 times larger than its actual size. In the case of the low power objective lens, its magnification is usually around 10x, meaning it magnifies the specimen by a factor of 10.
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