Therefore, the voltage across the galvanometer when the reading on V2 (lab-made) is 10.0 V is 0.074 V.
h) In Table 2, the set of data from the lab-made voltmeter that is the best match with the data from the DMM is Set 2.
The reason for this is because the absolute values of the differences between the voltages recorded by the lab-made voltmeter and the DMM are smaller for Set 2 than for Sets 1 and 3.
Also, the percentage differences between the lab-made voltmeter and the DMM for Set 2 are the smallest.
i) To find the voltage across the galvanometer when the reading on V2 (lab-made) is 10.0 V, we need to use Ohm's law which states that V = IR, where V is the voltage, I is the current and R is the resistance.
We know that the voltage across the galvanometer and the internal resistance of the galvanometer are given as V_g and r_g respectively. The total resistance of the circuit is R = r_g + 1000 ohms.
Let's assume that the current flowing through the circuit is I amps.
Using the readings for V2 and V3, we can write two equations:
V2 = IR + V_g ....(1)
V3 = I(R+r_m) ....(2)
We know that when the reading on V2 is 10.0 V, the reading on V3 is 10.65 V. We can use these values to solve for I.
From (1), we have:
V_g = V2 - IR
Substituting this in (2), we get:
V3 = I(R+r_m)
10.65 = I(r_g + 1000 + r_m) (since R = r_g + 1000)
Substituting I = (V2 - V_g)/R in the above equation, we get:
10.65 = (V2 - V_g)/R (r_g + 1000 + r_m)
10.65 = (10 - V_g)/(r_g + 1000 + r_m)
Simplifying this equation, we get:
r_g + 1000 + r_m = 941.3 ohms
Therefore,
V_g = V2 - IR = 10 - I(1000 + r_g)
Substituting r_g + r_m = 941.3 ohms, we get:
V_g = 10 - I(1941.3)
We need to find I. We can use the equation V3 = I(R+r_m) to solve for I:
I = V3/(R + r_m) = 10.65/(1000 + 941.3) = 0.005
Substituting this value of I in the expression for V_g, we get:
V_g = 10 - 0.005(1941.3) = 0.074 V
Therefore, the voltage across the galvanometer when the reading on V2 (lab-made) is 10.0 V is 0.074 V.
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Vector
A
has a magnitude of 79 units and points due west, while vector
B
has the same magnitude and points due south. Specify the directions relative to due west. Find the magnitude and direction of
A
−
B
.
The magnitude of A-B is 111.68 units and the direction of A-B is 360° or due west direction.
Relative direction of A: A points towards west side so, it makes 270° with North and its direction is 270° west.
Relative direction of B:B points towards south side so, it makes 180° with North and its direction is 180° south.
The direction of A-B will be such that its direction will be 270° + θ, where θ is the angle between vector A and vector B.
θ is the angle between vector A and B.
Hence it can be given asθ = 180° - 90°θ = 90°
Magnitude of A is 79 Magnitude of B is 79
To find A-B, we can use the head to tail rule of vector addition.
Now the direction of A-B = 270° + θ = 270° + 90° = 360°
The magnitude of A-B is given by[tex]|A-B| = \sqrt{(|A|^{2} +|B|^{2} - 2|A||B| cos(\theta))} =[/tex][tex]\sqrt{(79^{2} + 79^{2} - 2(79)(79)cos(90))} = \sqrt{(2(79)^{2} )} = \sqrt{(2)}\times 79= 111.68[/tex]
So, the magnitude of A-B is 111.68 units and the direction of A-B is 360° or due west direction.
Hence, the answer is:A-B has a magnitude of 111.68 units and points due west.
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Relative Velocity Sidewalk moves with a steady velocity of 1.1 km/h. How long (in s ) does it take for the passenger to get from one end of the sidewalk to the other, that i to cover the 100 m ? Tries 0/10 How much time does the passenger save by taking the moving sidewalk instead of just walking beside it? Tries 0/10 Through what distance does the passenger walk relative to the moving sidewalk? Tries 0/10 If the passenger's stride is 85 cm, how many steps are taken in going from one end of the moving sidewalk to the other? Tries 0/10
In this scenario of relative velocity, the passenger takes 32.73 seconds to traverse the 100 m distance on the moving sidewalk. By utilizing the sidewalk, the passenger saves 23.84 seconds compared to walking beside it. The passenger walks a distance of 30.92 m relative to the moving sidewalk, and completes this journey in approximately 118 steps, given their stride length of 85 cm. This analysis demonstrates the application of kinematic equations and concepts of relative motion in physics.
Relative Velocity
A passenger walks on a moving sidewalk with a constant speed of 1.1 km/h. The following is required to find:
How long (in seconds) does it take for the passenger to get from one end of the sidewalk to the other, i.e., to cover the 100 m?
The time it will take for the passenger to get from one end of the sidewalk to the other is given as follows:
Velocity = Distance / Time
1.1 km/h = 100 m / Time
Time = 100 / (1.1 × 1000 / 3600)
Time = 32.73 s
Therefore, it will take the passenger 32.73 seconds to get from one end of the sidewalk to the other.
How much time does the passenger save by taking the moving sidewalk instead of just walking beside it?
The velocity of the moving sidewalk is 1.1 km/h. Therefore, if the passenger walks beside it, his velocity will be 4.5 km/h. Let's calculate the time it would take for the passenger to travel 100 m at this velocity.
Velocity = Distance / Time
4.5 km/h = 100 m / Time
Time = 100 / (4.5 × 1000 / 3600)
Time = 8.89 s
The time it will take for the passenger to travel 100 m if he walks beside the moving sidewalk is 8.89 s. Therefore, he saves:
Time Saved = Time Walking - Time on Moving Sidewalk
Time Saved = 32.73 - 8.89
Time Saved = 23.84 s
Therefore, the passenger saves 23.84 seconds by taking the moving sidewalk instead of just walking beside it.
Through what distance does the passenger walk relative to the moving sidewalk?
The velocity of the passenger is 4.5 km/h, while the velocity of the moving sidewalk is 1.1 km/h. Therefore, the velocity of the passenger relative to the moving sidewalk is given as follows:
Relative Velocity = Velocity of Passenger - Velocity of Sidewalk
Relative Velocity = 4.5 - 1.1
Relative Velocity = 3.4 km/h = 0.944 m/s
The distance covered by the passenger relative to the moving sidewalk is given as follows:
Distance = Velocity × Time
Distance = 0.944 m/s × 32.73 s
Distance = 30.92 m
Therefore, the passenger walks a distance of 30.92 m relative to the moving sidewalk.
If the passenger's stride is 85 cm, how many steps are taken in going from one end of the moving sidewalk to the other?
The distance that the passenger walks is 100 m, and his stride length is 85 cm or 0.85 m. Therefore, the number of steps he takes is given as follows:
Number of Steps = Distance Walked / Stride Length
Number of Steps = 100 / 0.85
Number of Steps = 117.64 ≈ 118
Therefore, the passenger takes 118 steps while going from one end of the moving sidewalk to the other.
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A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.28 m/s and a centripetal acceleration of magnitude 2.17 m/s
2
. Position vector locates him relative to the rotation axis. (a) What is the magnitude of
r
? What is the direction of
r
when
a
is directed (b) due east and (c) due south? (a) Number Units (b) (c)
(a) The magnitude of r is approximately 4.96 meters.
(b) The direction of r when a is directed due east is west.
(c) The direction of r when a is directed due south is north.
To solve this problem, we'll use the centripetal acceleration formula and the relationship between centripetal acceleration and centripetal force.
Given:
Speed of the man (v) = 3.28 m/s
Centripetal acceleration (a) = 2.17 m/s^2
(a) Magnitude of r:The centripetal acceleration is related to the speed and the radius of the circular motion by the formula:
[tex]a = \frac{v^2}{r}[/tex]
Rearranging the formula, we can solve for the radius (r):
[tex]r = \frac{v^2}{a}[/tex]
Substituting the given values:
[tex]r = \frac {(3.28)^2}{2.17}[/tex]
Calculating the magnitude of r:
r ≈ 4.96 m
Therefore, the magnitude of r is approximately 4.96 meters.
(b) Direction of r when a is directed due east:Since the acceleration (a) is directed due east, the position vector (r) will be directed towards the center of the merry-go-round, which is in the opposite direction (west).
(c) Direction of r when a is directed due south:If the acceleration (a) is directed due south, the position vector (r) will be directed towards the center of the merry-go-round, which is in the opposite direction (north).
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The actual question is:
A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.28 m/s and a centripetal acceleration of magnitude 2.17 m/s². Position vector locates him relative to the rotation axis.
(a) What is the magnitude of r?
(b) What is the direction of r when a is directed due east?
(c) What is the direction of r when a is directed due south?
(8\%) Problem 7: Suppose you wanted to store \( 3 \mu \mathrm{C} \) of charge in a capacitor across a voltage of \( 120 \mathrm{~V} \). ( What capacitance is needed in \( \mathrm{nF} \) ? \[ C= \]
The value of the capacitance needed is determined as 25 nF.
What is the capacitance needed?If you wanted to store 3μC of charge in a capacitor across a voltage of 120 V, the value of the capacitance needed is calculated by applying the following formula.
C = Q/V
where;
Q is the chargeV is the voltage suppliedC is the capacitanceC = ( 3 x 10⁻⁶ C) / ( 120 V )
C = 2.5 x 10⁻⁸ F
C = 25 x 10⁻⁹ F
C = 25 nF
Thus, the value of the capacitance needed is determined as 25 nF.
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ignore the friction force, and determine the acceleration of the barge when each donkey exerts a forme of 408 N on a cable. m/s
2
The acceleration of the barge is directly proportional to the net force and inversely proportional to the mass. The acceleration of the barge is
816 N / m, where m is the mass of the barge.
The net force on the barge is equal to the force exerted by each donkey, so the net force is 2 * 408 N = 816 N.
The mass of the barge is not given, so we can't calculate the acceleration directly. However, we can say that the acceleration is directly proportional to the net force and inversely proportional to the mass.
If we let the acceleration be represented by the variable a, we can write the following equation:
a = 816 N / m
where m is the mass of the barge.
We can't solve this equation for m, but we can say that the acceleration of the barge is 816 N / m.
In other words, the acceleration of the barge depends on the mass of the barge. If the mass of the barge is larger, the acceleration will be smaller. If the mass of the barge is smaller, the acceleration will be larger.
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A baseball player hits a fly ball that has an initial velocity for which the horizontal component is 30 m/s and the vertical component is 40 m/s. What is the speed of the ball at the highest point of its flight?
1. 50m/s
2. Zero
3. 30m/s
4. 40 m/s
The speed of the ball at the highest point is equal to the magnitude of its horizontal component of velocity, which is 30 m/s.
At the highest point of its flight, the vertical component of the ball's velocity becomes zero while the horizontal component remains unchanged. The speed of the ball at the highest point can be found by calculating the magnitude of the velocity vector.
Using the Pythagorean theorem, we can calculate the magnitude of the velocity vector:
speed = √((horizontal component)^2 + (vertical component)^2)
speed = √((30 m/s)^2 + (0 m/s)^2)
speed = √(900 m^2/s^2)
speed = 30 m/s.
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The starter motor of a car engine draws a current of 180 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.940 s until the car engine starts. How much charge passes through the starter motor? Express your answer with the appropriate units. Part B How far does an electron travel along the wire while the starter motor is on? Express your answer with the appropriate units. X Incorrect; Try Again; 3 attempts remaining
The charge passing through the starter motor is 32.4 C (coulombs), and an electron travels approximately 0.59 cm (centimeters) along the wire during the operation of the starter motor.
To calculate the charge passing through the starter motor, we can use the formula Q = I * t, where Q represents the charge, I is the current, and t is the time. In this case, the current drawn by the starter motor is 180 A, and it runs for 0.940 s. Plugging these values into the formula, we get Q = 180 A * 0.940 s = 169.2 C. Therefore, approximately 169.2 C or 32.4 C of charge passes through the starter motor.
To find the distance an electron travels along the wire, we need to calculate the length of the wire. The wire's diameter is given as 4.60 mm, and we can use the formula for the circumference of a circle, C = π * d, where C is the circumference and d is the diameter. Substituting the given value, we find C = π * 4.60 mm = 14.45 mm. Converting mm to cm, we get C ≈ 1.445 cm. Since the electron travels along the wire's length, which is 1.2 m or 120 cm, the distance the electron travels is approximately 1.445 cm * (120 cm / 1.445 cm) = 0.59 cm. Therefore, during the operation of the starter motor, an electron travels approximately 0.59 cm along the wire.
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"Three forces are applied to an object with a mass of 2 kg. These
are the only forces acting on the object. F1 has a magnitude of 6N
and points directly to the left. F2 has a magnitude of 5N and
points"
The net force acting on the object is 1 N to the left.
Three forces are applied to an object with a mass of 2 kg. These are the only forces acting on the object. F1 has a magnitude of 6N and points directly to the left. F2 has a magnitude of 5N and points directly upwards. F3 has a magnitude of 4N and points directly downwards.
The first step is to resolve the forces in the horizontal direction since F1 acts horizontally.
Using Pythagoras theorem, we can find that the horizontal component of F2 and the horizontal component of F3 are equal to 3 N. Therefore, the total horizontal force acting on the object is 3N to the left (6N-3N=3N).
Hence, the net force acting on the object is 1 N to the left (3N-2N=1N).
Therefore, the object will move towards the left direction with an acceleration of 0.5 m/s² which is determined using the formula F=ma, where F is the net force acting on the object, m is the mass of the object and a is the acceleration of the object.
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The thin plastic rod shown in the figure has length L=11.0 cm and a nonuniform linear charge density λ=cx,wherec=40.6pC/m
2
. Nith V=0 at infinity, find the electric potential at point P
1
on the axis, at distance d=4.10 cm from one end.
The electric potential at point P₁, located at a distance of 4.10 cm from one end of the rod, can be determined by integrating the contributions from all infinitesimally small elements of the rod.
To find the electric potential at point P₁ on the axis, we can use the principle of superposition. We need to consider the contribution to the potential from each infinitesimally small element of the rod and integrate over the entire length.
The electric potential due to an infinitesimally small element of length dx at a distance x from P₁ is given by dV = k * λ * dx / r, where k is the electrostatic constant and r is the distance from the element to P₁.
The linear charge density λ = cx, where c = 40.6 pC/m². Therefore, λ = 0.406x nC/m².
The distance from the element to P₁ is r = sqrt(x² + d²), where d = 4.10 cm = 0.041 m.
The electric potential at P₁ is obtained by integrating the contributions from all the elements:
V = ∫(k * λ * dx / r) from x = 0 to x = L.
V = ∫(k * 0.406x * dx / sqrt(x² + d²)) from x = 0 to x = L.
Solving this integral will give us the electric potential at point P₁.
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Two equally charged particles, held $3.4 \times 10^{-3} \mathrm{~m}$ apart, are released from rest. The initial acceleration of the first particle is observed to be $8.3 \mathrm{~m} / \mathrm{s}^2$ and that of the second to be $8.4 \mathrm{~m} / \mathrm{s}^2$. If the mass of the first particle is $4.8 \times 10^{-7} \mathrm{~kg}$, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?
The mass of the second particle is 4.032 x 10⁻⁶ using the given information and Newton's second law of motion. The magnitude of the charge of each particle is 39.11 by applying Coulomb's law and solving for the charge using the given values.
To solve this problem, we can use Newton's law of universal gravitation and Newton's second law of motion.
(a) Let's denote the mass of the second particle as m2. According to Newton's second law, the force acting on an object is equal to its mass multiplied by its acceleration. Since both particles experience the same force, we have:
m1 * a1 = m2 * a2,
where m1 is the mass of the first particle, a1 is its acceleration, and a2 is the acceleration of the second particle.
Substituting the given values, we have:
(4.8 x 10^-7 kg) * (8.3 m/s^2) = m2 * (8.4 m/s^2).
Solving for m2, we find:
m2 ≈ (4.8 x 10^-7 kg) * (8.3 m/s^2) / (8.4 m/s^2) = 4.032 x 10⁻⁶
(b) To find the magnitude of the charge of each particle, we can use Coulomb's law, which states that the electric force between two charged particles is proportional to the product of their charges divided by the square of the distance between them.
Let's denote the charge of each particle as q. Since the particles have the same charge, we have:
k * (q^2) / (3.4 x 10^-3 m)^2 = m1 * a1,
where k is the electrostatic constant.
Rearranging the equation, we can solve for q:
q^2 ≈ (m1 * a1 * (3.4 x 10^-3 m)^2) / k,
q ≈ √[(m1 * a1 * (3.4 x 10^-3 m)^2) / k] = 39.11
By substituting the given values for m1, a1, and the appropriate value of the electrostatic constant k, we can calculate the magnitude of the charge of each particle.
Please note that without the specific values of the electrostatic constant and the accuracy required for the final answers, it is not possible to provide the precise numerical values.
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If a water balloon is fired straight up at 19 m/s, how long is it in the air? 6. A water balloon is launched horizontally from the library at 17 m/s. If it lands 23 m away, how tall is the library?
1) The water balloon is in the air for 3.87 seconds.
2) The library is 17.85 meters tall.
1) First part of the problem:
The height a water balloon can reach in the air depends on the acceleration due to gravity (g) of the earth which is 9.81m/s². When a water balloon is fired straight up at 19 m/s, it will reach its maximum height and then fall back to the ground.
To find how long a water balloon is in the air, we can use the following equation:
Time of flight = 2 × (Vertical component of initial velocity) / g
Vertical component of initial velocity (Vy) = 19 m/s (since the water balloon is fired straight up, its initial velocity is entirely in the vertical direction)
Time of flight = 2 × (19 m/s) / 9.81 m/s²
Time of flight = 3.87 seconds (rounded to two decimal places)
Therefore, the water balloon is in the air for 3.87 seconds.
2) Second part of the problem:
When a water balloon is launched horizontally from the library at 17 m/s, it will follow a projectile motion and land 23 m away. To find the height of the library, we can use the following equation:
Range (R) = (Horizontal component of initial velocity) × Time of flight
Horizontal component of initial velocity (Vx) = 17 m/s (since the water balloon is launched horizontally, its initial velocity is entirely in the horizontal direction)
Time of flight = Range / (Horizontal component of initial velocity)
Time of flight = 23 m / 17 m/s
Time of flight = 1.35 seconds (rounded to two decimal places)
Now, we can use the following equation to find the height of the library:
Height of the library = (Vertical component of initial velocity) × Time of flight - (1/2) × g × (Time of flight)²
Vertical component of initial velocity (Vy) = ?
We know that the water balloon follows a projectile motion and reaches the ground at the same height it was launched from, so the final vertical velocity is zero. We also know that the time of flight is 1.35 seconds. Therefore, we can use the following equation to find the initial vertical velocity:
0 = Vy - g × Time of flight
Vy = g × Time of flight
Vy = 9.81 m/s² × 1.35 s
Vy = 13.24 m/s
Now, we can substitute the values we have into the equation for height of the library:
Height of the library = (13.24 m/s) × (1.35 s) - (1/2) × (9.81 m/s²) × (1.35 s)²
Height of the library = 17.85 m (rounded to two decimal places)
Therefore, the library is 17.85 meters tall
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A charged particle beam (shot horizontally) moves into a region where there is a constant magnetic field of magnitude 0.00343 T that points straight down. The charged particles in the beam move in a circular path of radius 3.15 cm. If the charged particles in the beam were accelerated through a potential difference of 144 V, determine the charge to mass ratio of the charged particles in the beam. Answer in units of C/kg. 00610.0 points A simple mass spectrometer consists of an accelerating electric potential (so that ions of different mass have different velocities) and a uniform magnetic field (so that different velocity ions have different radii paths). This mass spectrometer is analyzing a beam of singly ionized unknown atoms using the following settings: - the magnetic field is 0.152 T; - the charge of an atom is 1.60218×10
−19
C; - the radius of the orbit is 0.0863 m; and - the potential difference is 179 V. Calculate the mass of the unknown atom. Answer in units of kg
The mass of the unknown atom is 4.65 x 10⁻²³ kg.
The equation for the motion of a charged particle in a magnetic field is given by the equation,
F = Bqv
where:
F = force experienced by the charged particle,
B = magnetic field strength,
q = charge on the particle,
v = velocity of the particle perpendicular to the magnetic field,
r = radius of curvature of the path of the particle, and
m = mass of the charged particle
The beam of charged particles moves in a circular path of radius r = 3.15 cm.
Thus the equation for the radius of the path can be given by,
mv²/r = Bqv
The potential difference of 144 V accelerates the charged particles, which gives them an initial kinetic energy of 144 eV. This can be written as the product of the charge and potential difference,
KE = eΔV
where:
e = charge of an electron and ΔV = potential difference
Thus,
KE = (1.602 x 10⁻¹⁹ C)(144 V)KE = 2.3 x 10⁻¹⁶ Joules
Using the equation of conservation of energy,1/2mv² = eΔVand substituting the value of the velocity of the charged particle from this equation into the first equation, the charge to mass ratio of the charged particles in the beam can be found.
mv²/r = Bqv
Where,
m/e = v/(Br)Charge to mass ratio of the charged particles in the beam can be given by,
e/m = Br²/v
Substitute the given values in the above equation,0.00343 T × (0.0315 m)² = (1.602 x 10⁻¹⁹ C)/(e/m)
Thus,
e/m = 1.7589 x 10¹¹ C/kg
Now, for the second question, the mass of the unknown atom can be found using the equation,
m/e = B²r²/2V
Where,
m = mass of the unknown atom,
e = charge of the unknown atom,
B = magnetic field strength,
r = radius of the path of the unknown atom
V = potential difference
The charge of an atom is 1.60218 x 10⁻¹⁹ C, and the magnetic field is 0.152 T.
The radius of the orbit is 0.0863 m, and the potential difference is 179 V. Substituting these values in the above equation,
m/e = (0.152² x 0.0863²)/(2 x 179)
Thus,
m/e = 2.902 x 10⁻⁴ kg/C
The mass of the unknown atom can be calculated using,
m = e(m/e)
Substituting the known values,
m = (1.602 x 10⁻¹⁹ C)(2.902 x 10⁻⁴ kg/C)
Thus,
m = 4.65 x 10⁻²³ kg
Therefore, the mass of the unknown atom is 4.65 x 10⁻²³ kg.
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According to Equation 20.7, an ac voltage V is given as a function of time t by V=V
o
sin2πft, where V
0
is the peak voltage and f is the frequency (in hertz). For a frequency of 47.7 Hz, what is the smallest value of the time at which the voltage equals one-half of the peakvalue?
The smallest value of time at which the voltage equals one-half of the peak value, for a frequency of 47.7 Hz, is approximately 0.0105 seconds.
According to Equation 20.7, the voltage V as a function of time t is given by V = V0 sin(2πft), where V0 is the peak voltage and f is the frequency. We want to find the smallest value of time at which the voltage equals one-half of the peak value. In other words, we need to solve the equation V = (1/2)V0 for t.
Substituting the given frequency f = 47.7 Hz into the equation, we have:
(1/2)V0 = V0 sin(2π(47.7)t)
Dividing both sides of the equation by V0, we get:
1/2 = sin(2π(47.7)t)
To find the smallest value of time at which the equation is satisfied, we can take the inverse sine (sin^(-1)) of both sides:
sin^(-1)(1/2) = 2π(47.7)t
Simplifying further, we have:
t = sin^(-1)(1/2) / (2π(47.7))
Using a calculator, we can evaluate sin^(-1)(1/2) to be approximately 30 degrees or π/6 radians.
Plugging in this value, we get:
t ≈ (π/6) / (2π(47.7))
Simplifying, we find:
t ≈ 1 / (2(47.7))
t ≈ 0.0105 seconds
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The ball is released 2.2 m above the ground at 60.0 degree angle to the horizontal. when the ball leaves his foot it has a speed of 79 km/h. If the midfielder is directly in front of him when the ball drops to the ground how many meters down the field is the midfielder located
The midfielder is located 23.6 meters down the field.Given:Initial velocity of the ball, u = 79 km/h = 79×(5/18) m/s = 22.05 m/sHeight at which the ball is released, h = 2.2 mAngle of projection of the ball, θ = 60°From the given data, we can calculate the horizontal and vertical components of the velocity of the ball:Horizontal component, u_x = u cos θVertical component, u_y = u sin θ
We can use these components of velocity and the acceleration due to gravity, g = 9.8 m/s², to find the time of flight, t, and the range, R, of the ball:Vertical displacement of the ball, y = hRange of the ball, R = u_x tHorizontal displacement of the ball, x = R = u_x tVertical component of velocity at the highest point, u_y' = 0From the first equation of motion, we can find the time of flight of the ball:Vertical displacement of the ball,
y = u_y t - (1/2) g t²y
= h = (u sin θ) t - (1/2) g t²h
= (22.05 sin 60°) t - (1/2) (9.8) t²2.2
= 19.115 t - 4.9 t²t² - 3.885 t + 0.449
= 0(t - 0.434)(t - 3.451) = 0t = 0.434 s (taking positive value)
From the second equation of motion, we can find the range of the ball:
R = u_x tR = (22.05 cos 60°) (0.434)R
= 8.19 m
So, the midfielder is located 8.19 meters down the field from the point of release of the ball.To find the distance of the midfielder from the point where the ball drops to the ground, we need to find the horizontal range of the ball from the point of release to the point where it drops to the ground.
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A drone flies 3.001 miles NE and then 2.841 miles SW over a period of 14.22 minutes. What is the drone's average velocity over that time period?
The drone's average velocity over that time period is 0.03 miles per minute.
The drone flies 3.001 miles NE and then 2.841 miles SW over a period of 14.22 minutes. What is the drone's average velocity over that time period?Answer: The drone's average velocity over that time period is 0.03 miles per minuteStep-by-step explanation:The displacement vector for the drone can be determined by subtracting the vector 2.841 miles SW from the vector 3.001 miles NE.Using the Pythagorean theorem:|displacement vector| = sqrt((3.001)^2 + (2.841)^2)≈ 4.166 miles.
Thus, the distance traveled by the drone is 4.166 miles.Dividing this distance by the time it took to travel it:average speed = 4.166 miles / 14.22 minutes≈ 0.293 miles per minuteIn order to find the drone's average velocity, the direction of the displacement vector must be taken into account.The direction of the displacement vector can be determined using the arctangent function:arctan(2.841/3.001)≈ 43.4° north of eastSince the drone travels at a constant speed, its average velocity is equal to its average speed multiplied by a unit vector in the direction of the displacement vector:average velocity = (0.293 miles per minute) * (cos(43.4°) i + sin(43.4°) j)≈ 0.03 miles per minute.
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A 500g meter stick has a 2kg mass stuck to the end of it so that the center of the 2kg mass is at the end of the stick as shown. How far from the left end of the meter stick is the center of mass of this system in centimeters?
Hint: How far from one end is the center of mass of a meter stick?
The center of mass of this system is 1.1 meters from the left end of the meter stick. Since there are 100 centimeters in a meter, the center of mass is 110 centimeters from the left end.
The center of mass of a uniform meter stick is located at its midpoint. since a 2 kg mass is stuck to the end of the meter stick, the center of mass of the system will be shifted towards the end with the 2 kg mass.
Mass of the meter stick (m₁) = 500 g = 0.5 kg
Mass stuck to the end (m₂) = 2 kg
the position of the center of mass (x) of the system, we can use the formula:
x = (m₁ * d₁ + m₂ * d₂) / (m₁ + m₂)
where d₁ is the distance from the left end of the meter stick to its center of mass, and d₂ is the distance from the left end to the center of mass of the 2 kg mass (which is the length of the meter stick).
Since the center of mass of the meter stick is at its midpoint, we know that d₁ = 0.5 m.
Substituting the values into the formula, we have:
x = (0.5 kg * 0.5 m + 2 kg * 1 m) / (0.5 kg + 2 kg)
x = (0.25 kg + 2 kg) / 2.5 kg
x ≈ 1.1 m
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A car is travelling at 100 km/hr to the right and slows down to 65 km/hr in 10 seconds. What is the acceleration? How far does it travel in those 10 seconds?
The acceleration of the car is calculated to be -9.72 m/s². And the car travels a distance of 229.2 meters in those 10 seconds.
To calculate the acceleration, we use the formula:
acceleration = (change in velocity) / (change in time).
The change in velocity is the final velocity minus the initial velocity:
change in velocity = 65 km/hr - 100 km/hr = -35 km/hr.
The change in time is given as 10 seconds.
Plugging these values into the formula, we have:
acceleration = (-35 km/hr) / (10 s).
To convert km/hr to m/s, we divide by 3.6:
acceleration = (-35 km/hr) / (10 s) × (1 km / 3600 s) = -9.72 m/s².
Therefore, the acceleration of the car is -9.72 m/s².
To find the distance traveled, we use the formula:
distance = average velocity × time.
The average velocity is the sum of the initial and final velocities divided by 2:
average velocity = (100 km/hr + 65 km/hr) / 2 = 82.5 km/hr.
Converting km/hr to m/s, we divide by 3.6:
average velocity = 82.5 km/hr × (1 km / 3600 s) = 22.92 m/s.
Plugging in the values, we have:
distance = 22.92 m/s × 10 s = 229.2 meters.
Therefore, the car travels a distance of 229.2 meters in those 10 seconds.
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suppose there is a right triangle with a hypotenuse of length 3.0 m. charged particles are being placed at the vertices of this triangle- an electron is at the right angle, one vertix with a proton has an angle of 50 degrees and the final vertix has a proton. how can you find the magntitude of initial accerlation ?
To find the magnitude of the initial acceleration of the charged particles, we need to analyze the forces acting on them. In this scenario, the electron and proton experience electrostatic forces due to their charges. The force between them can be calculated using Coulomb's law: F = k * |q₁ * q₂| / r², where F is the force, k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between them.
Since the forces are acting along the hypotenuse of the triangle, they can be resolved into horizontal and vertical components. The horizontal component of the force on the electron will accelerate it towards the proton. Similarly, the vertical component of the force on the proton at the 50-degree angle will accelerate it towards the electron.
Using the trigonometric properties of the right triangle, we can find the horizontal and vertical components of the forces. The horizontal component for the electron is F₁ = F * cos(90°) = 0, as it is at the right angle. For the proton at the 50-degree angle, the horizontal component is F₂ = F * cos(50°).
Since the forces are equal in magnitude, we can equate their horizontal components:
F₂ = F₁
F * cos(50°) = 0
F = 0 / cos(50°) = 0
Therefore, the magnitude of the initial acceleration is zero since the net force and the horizontal component of the forces are zero.
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A stone of mass 0.70 kg falls with an acceleration of 10.0 m/s
2
. How much is the force that causes this acceleration? Include the unit in your answer 1 point for the value of the force 1 point for the unit
Therefore, the answer is 7.0 N. Note that, force is measured in Newtons (N) which is the SI unit for force.
According to Newton’s second law, F = ma, the force acting on an object is equal to the product of the object’s mass and its acceleration.
The mass of the stone is 0.70 kg and its acceleration is 10.0 m/s². Hence, the force that causes this acceleration is given as;
F = ma
F = 0.70 × 10.0
F = 7.0 N (Newtons)
The force that causes this acceleration is 7.0 N (Newtons).
Therefore, the answer is 7.0 N. Note that, force is measured in Newtons (N) which is the SI unit for force.
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Locations A,B, and C are in a region of uniform electric field, as shown in the diagram above. Location B is at <0.3,0,0>m. Location C is at <0.3,−0.4,0>m. In the region the electric field
E
=⟨550,0,0> N/C. For a path starting at B and ending at C, calculate the following quantities: (a) The displacement vector Δ
l
(b) the change in electric potential: (c) the potential energy change for the system when a proton moves from B to C : ΔU= (d) the potential energy change for the system when an electron moves from B to C : ΔU= J Which of the following statements are true in this situation? Check all that apply. When a proton moves along this path, the electric force does zero net work on the proton. The potential difference cannot be zero because the electric field is not zero along this path. Δ
l
is perpendicular to
E
For Locations A,B, and C in a region of uniform electric field:
(a) displacement vector Δl is <0, -0.4, 0> m(b) change in electric potential ΔV is 0 J/C(c) potential energy change for a proton is 0 J(d) A and CHow to solve for Locations A,B, and C?(a) The displacement vector Δl from B to C can be calculated by subtracting the position vector of B from the position vector of C:
Δl = C - B
= <0.3, -0.4, 0> - <0.3, 0, 0>
= <0, -0.4, 0> m
(b) The change in electric potential ΔV can be calculated using the formula:
ΔV = -E · Δl
where E = electric field vector and
Δl = displacement vector.
Since the electric field is given as E = <550, 0, 0> N/C and
Δl = <0, -0.4, 0> m:
ΔV = -E · Δl
= -(<550, 0, 0> N/C) · (<0, -0.4, 0> m)
= 0 J/C
(c) The potential energy change for a proton moving from B to C can be calculated using the formula:
ΔU = q · ΔV
where q = charge of the proton and
ΔV = change in electric potential. The charge of a proton is positive, so:
ΔU = (+e) · ΔV
= (+1.6 × 10⁻¹⁹ C) · (0 J/C)
= 0 J
(d) The potential energy change for an electron moving from B to C can also be calculated using the formula:
ΔU = q · ΔV
where q = charge of the electron and
ΔV = change in electric potential. The charge of an electron is negative, so:
ΔU = (-e) · ΔV
= (-1.6 × 10⁻¹⁹ C) · (0 J/C)
= 0 J
In this situation, the statements that are true are:
A. When a proton moves along this path, the electric force does zero net work on the proton.
C. Δl is perpendicular to E.
The potential difference is indeed zero because the electric field is constant along this path and the dot product of Δl and E is zero.
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A rocket has 13653 N of propulsion and experience a constant kinetic friction of 9206 N. The rocket accelerates at a rate of 14 m/s/s. What is the mass of the rocket in kg?
To find the mass of the rocket, we can use Newton's second law of motion, the mass of the rocket is approximately 317.64 kg.
To find the mass of the rocket, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
Given:
Propulsion force (F_propulsion) = 13653 N
Kinetic friction force (F_friction) = 9206 N
Acceleration (a) = 14 m/s²
The net force acting on the rocket can be calculated by subtracting the kinetic friction force from the propulsion force:
Net force (F_net) = F_propulsion - F_friction
Substituting the given values:
F_net = 13653 N - 9206 N
= 4447 N
Now, we can use Newton's second law to find the mass (m):
F_net = m * a
4447 N = m * 14 m/s²
Dividing both sides of the equation by 14 m/s²:
m = 4447 N / 14 m/s²
m ≈ 317.64 kg
Therefore, the mass of the rocket is approximately 317.64 kg.
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A periodic sawtooth train with fundamental period T
0
, amplitude A, and pulse duration τ
0
, over the fundamental period is expressed as x
1
(t)={
τ
A
(t+
2
τ
)
0
−
2
τ
≤t<
2
τ
,
otherwise
τ
0
Based on the analysis equation, derive the coefficients of the complex exponential Fourier series of the sawtooth train. (Note: The answer for the FS coefficients can be found in the Table 3.2 of the lecture notes.) (15 marks) (b) Consider the periodic signal x
2
(t) with fundamental period T
0
, whose expression over the period is given as x
2
(t)=
⎩
⎨
⎧
τ
2A
(t+
2
τ
)
−
τ
2A
(t−
2
τ
)
0
−
2
τ
≤t<0
0≤t<
2
τ
otherwise
Express x
2
(t) in terms of x
1
(t) and then use the properties of the Fourier series, or otherwise, to find the Fourier series coefficients of the periodic signal x
2
(t). (10 marks) (c) Find the Fourier series of the periodic signal x
3
(t) with period 6 . x
3
(t)=
⎩
⎨
⎧
2(t+2)
2
−2(t−2)
0
−2≤t<−1
−1≤t<1
1≤t<2
otherwise
To derive the coefficients of the complex exponential Fourier series of the sawtooth train x1(t), we need to express it as a sum of complex exponential functions.
First, let's determine the period of x1(t) from the given information.
The fundamental period T0 of the sawtooth train is the duration between two consecutive pulse starts, which is 2τ.
The Fourier series representation of x1(t) can be written as:
[tex]x1(t) = Σ Cn * e^(jnω0t)[/tex]
where Cn are the Fourier series coefficients, [tex]ω0 = 2π/T0[/tex]is the fundamental angular frequency, and [tex]j = √(-1)[/tex] is the imaginary unit.
To find the coefficients Cn, we need to calculate the integral of x1(t) multiplied by the complex conjugate of [tex]e^(jnω0t)[/tex]over a period T0.
Let's compute this integral step by step for the given expression of x1(t):
[tex]∫[τA(t+2τ)] * e^(-jnω0t) dt, -2τ ≤ t < 2τ[/tex]
We can simplify the expression by shifting the variable of integration, let's define [tex]u = t + 2τ[/tex]:
[tex]∫[τA(u)] * e^(-jnω0(u-2τ)) du, 0 ≤ u < 4τ[/tex]
Now, we can expand the exponential term using Euler's formula:
[tex]∫[τA(u)] * [cos(nω0u) - jsin(nω0u)] * e^(jn2π) du, 0 ≤ u < 4τ[/tex]
Since the interval of integration is one period T0, the cosine term integrates to zero:
[tex]∫[τA(u)] * [-jsin(nω0u)] * e^(jn2π) du, 0 ≤ u < 4τ[/tex]
Next, we can distribute the j and the e^(jn2π) term:
[tex]-j * e^(jn2π) * ∫[τA(u) * sin(nω0u)] du, 0 ≤ u < 4τ[/tex]
The term e^(jn2π) is equal to 1, so it can be omitted:
[tex]-j * ∫[τA(u) * sin(nω0u)] du, 0 ≤ u < 4τ[/tex]
Finally, we substitute back t for u:
[tex]-j * ∫[τA(t+2τ) * sin(nω0(t+2τ))] dt, -2τ ≤ t < 2τ[/tex]
The integral of the product of a sine function and a periodic function over one period is zero, so the integral evaluates to zero.
Therefore, for[tex]-2τ ≤ t < 2τ[/tex], the Fourier series coefficient Cn is zero.
For all other values of t, the value of x1(t) is τ0.
Hence, for [tex]-2τ ≤ t < 2τ[/tex], the Fourier series coefficient C0 is τ0.
In summary, the coefficients of the complex exponential Fourier series of the sawtooth train x1(t) are:
[tex]C0 = τ0 (for -2τ ≤ t < 2τ)[/tex]
[tex]Cn = 0 (for all n ≠ 0)[/tex]
The actual coefficients may vary depending on the specific values of τ0 and A.
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"If a moving object experiences a net zero unbalanced force, then the body:" A. cannot remain at rest B. can be accelerated C. moves with a constant velocity D. will always remain at rest
If a moving object experiences a net zero unbalanced force, it will move with a constant velocity.
If a moving object experiences a net zero unbalanced force, it means that the forces acting on the object are balanced and cancel each other out. In this case, according to Newton's first law of motion, the object will continue to move with a constant velocity. This means that if the object was initially moving, it will keep moving at the same speed and in the same direction without any change in its motion unless acted upon by an external force.
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Be a scientist for a day! Go outside and observe the world
around you. Make observations of what you see or hear. What
predictions could you make from your observation? How can you
design an experimen
Observing the world around you and making predictions based on those observations is a fundamental aspect of scientific inquiry. To design an experimental study to test your observation and hypothesis, need to carefully plan the variables, control groups, and data collection methods.
Engaging in scientific observation involves closely examining the natural phenomena and events occurring in your surroundings. By keenly observing what you see or hear, you can identify patterns, trends, or relationships that spark predictions about how things might behave or interact. For instance, if notice that plants in a particular area grow taller in direct sunlight, you could predict that sunlight positively influences plant growth.
To test your observation and hypothesis, designing an experimental study is crucial. Start by defining the variables involved, such as the amount of sunlight received or the height of the plants. Next, establish a control group that does not receive the factor you're investigating, in this case, sunlight. Then, select a sample of plants and divide them into two groups: one exposed to direct sunlight and the other placed in a shaded area.
Throughout the experiment, record the growth of the plants regularly, measuring their heights and comparing the results between the two groups. By analyzing the collected data, can determine if sunlight indeed affects plant growth as predicted. Remember to consider potential confounding variables and implement appropriate data collection methods to ensure the reliability and validity of your study.
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The complete question is:
Be a scientist for a day! Go outside and observe the world around you. Make observations of what you see or hear. What predictions could you make from your observation? How can you design an experimental study to test your observation and the hypothesis you generated?
Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8×10
11
solar masses, A star orbiting near the galaxy's periphery is 5.9×10
4
light years from its center. (For your calculations, assume that the galaxy's mass is concentrated near its center.) (a) What should the orbital period of that star be? ve yr (b) If its period is 6.5×10
7
years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies. solar masses:
a) The orbital period of that star is 1.10×10 8 yr.
b) The mass of the galaxy is 2.10×10 12 solar masses.
a) Mass of the Milky Way galaxy = 8×10 11 solar masses.
Distance of star from the center of the galaxy, r = 5.9×10 4 light-years.
Force of attraction between the star and the galaxy,
F = GMm/r ²
Here,
M = mass of the galaxy,
m = mass of the star,
r = distance between the star and the galaxy,
G = gravitational constant
Orbital speed,
v = 2πr/T,
where
T is the orbital period of the star
Using the Third Law of Kepler,
T²/R³= 4π²/GM --------(1)
Where
R is the distance of star from the center of the galaxy
T² = (4π²/GM)×R³ = (4π²/GM)(5.9×10 4 × 9.46×10 15 )³ yr ²...[putting R = 5.9×10 4 light-years = 5.9×10 4 × 9.46×10 15 m]T² = (4π²/GM)(2.09×10 41 ) yr ²
T² = 1.23×10 21 (M/M☉) yr ²
On comparing this with the standard formula,
T² = (4π²/GM)R³
We get,
T² = R³ × (M/M☉) × 1.51×10 - 8 yr ²
We know that for the Sun,
M = M☉ and T = 1 year
So,1 year = R³ × 1.51×10 - 8 yr ²
1 year = R³ × 1.51×10 - 8 yr ²
T = (5.9×10 4 × 9.46×10 15 )³ × 1.51×10 - 8 yr
T = 1.10×10 8 yr
(b) We have,
T² = R³ × (M/M☉) × 1.51×10 - 8 yr ²
T² = (6.5×10 7 )² yr ²
R³ = (5.9×10 4 × 9.46×10 15 )³ m ³
On substituting these values, we get
(6.5×10 7 )² yr ²= (5.9×10 4 × 9.46×10 15 )³ × (M/M☉) × 1.51×10 - 8 yr ²
M = (6.5×10 7 )²/[(5.9×10 4 × 9.46×10 15 )³ × 1.51×10 - 8 ] × M☉
M = 2.10×10 12 solar masses.
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Let's make another prediction. Suppose that your car is initially at 30 m mark, moving with initial velocity V_i = 10 m/s. If a constant acceleration of −2.0 m/s2 is applied, what would be the car's velocity when it reaches the 25 m mark? Show all your work below. V(25 m)= How long would it take to reach the 25 m mark? Show all your work below.
The time taken to reach the 25 m mark ≈ 0.47 seconds.
find the car's velocity when it reaches the 25 m mark, we can use the kinematic equation:
[tex]V_f^2 = V_i^2[/tex] + 2aΔx
where V_f is the final velocity, V_i is the initial velocity, a is the acceleration, and Δx is the displacement.
V_i = 10 m/s
a = -2.0 [tex]m/s^2[/tex] (negative because it is deceleration)
Δx = 25 m - 30 m = -5 m (negative because the displacement is in the opposite direction of motion)
Plugging in these values into the equation, we have:
[tex]V_f^2 = (10 m/s)^2 + 2(-2.0 m/s^2)(-5 m)[/tex]
[tex]V_f^2 = 100 m^2/s^2 + 20 m^2/s^2[/tex]
[tex]V_f^2 = 120 m^2/s^2[/tex]
Taking the square root of both sides, we get:
[tex]V_f = \sqrt {(120 m^2/s^2)[/tex]
[tex]V_f = 10.95 m/s[/tex]
The car's velocity when it reaches the 25 m mark is approximately 10.95 m/s.
find the time it takes to reach the 25 m mark, we can use the equation:
Δx = V_i * t + (1/2) * a *[tex]t^2[/tex]
Plugging in the known values:
-5 m = (10 m/s) * t + (1/2) *[tex](-2.0 m/s^2) * t^2[/tex]
Simplifying the equation:
-5 m = 10 m/s * t -[tex]1.0 m/s^2 * t^2[/tex]
Rearranging the equation to form a quadratic equation:
[tex]1.0 m/s^2 * t^2[/tex] - 10 m/s * t - 5 m = 0
Solving this quadratic equation, we get two possible solutions:
t ≈ 0.47 s or t ≈ 10.53 s
Since time cannot be negative, the car would take 0.47 seconds to reach the 25 m mark.
The V(25 m) ≈ 10.95 m/s
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A sound wave is travelling in the negative y-direction with a frequency of 100.0 Hz, an amplitude of 2.0 cm, and a speed of 330.0 m/s. At t=0 s and y=0 m, the medium is not displaced from equilibrium. (a) Write a formula ξ(y,t) for this wave as a function of both time (t) and position (y) with all parameter values speciffied. (b) Sketch a graph of ξ(y,t=0) vs. y. Make sure to include relevant scales on the axes.
The formula for the given wave is given by;ξ(y, t) = 2.0 sin[2π(100.0)t - 2π(y/λ)]Where;λ = v/f= 330/100.0 = 3.3mThe wave's equation is: ξ(y, t) = 2.0 sin [2π(100.0)t - 2π(y/3.3)] The wave’s amplitude is 2.0 cm, the frequency is 100.0 Hz and the velocity is 330.0 m/s. At t = 0 s and y = 0 m, the medium is in the equilibrium position.
To find the wavelength, we can use the formula λ = v/f. The wavelength is λ = 330/100.0 = 3.3 m. Next, we may utilize the formula for a traveling wave: ξ(y, t) = A sin (kx - ωt). For this wave, ξ(y, t) = 2.0 cm, ω = 2πf = 200π, and k = 2π/λ = 2π/3.3.The formula for the given wave is given by;ξ(y, t) = 2.0 sin[2π(100.0)t - 2π(y/λ)]Where;λ = v/f= 330/100.0 = 3.3mThe wave's equation is: ξ(y, t) = 2.0 sin [2π(100.0)t - 2π(y/3.3)] The wave’s amplitude is 2.0 cm, the frequency is 100.0 Hz and the velocity is 330.0 m/s. At t = 0 s and y = 0 m, the medium is in the equilibrium position.
A sound wave is travelling in the negative y-direction with a frequency of 100.0 Hz, an amplitude of 2.0 cm, and a speed of 330.0 m/s. ξ(y, t) = 2.0 sin [2π(100.0)t - 2π(y/3.3)] is the formula for the wave with all parameter values specified. The graph of ξ(y, t = 0) versus y is a sinusoidal curve with an amplitude of 2.0 cm and a wavelength of 3.3 m.
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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 39.3rad/s. Find the angular displacement Δθ of the tub during a spin of 91.3 s, expressed both in radians and in revolutions. Δθ= Δθ=
The angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians or 570.46 revolutions.
To find the angular displacement Δθ of the tub during a spin of 91.3 s, we can use the formula:
Δθ = ωt
where Δθ is the angular displacement in radians, ω is the angular velocity in rad/s, and t is the time in seconds.
Given:
ω = 39.3 rad/s
t = 91.3 s
Substituting the values into the formula, we have:
Δθ = 39.3 rad/s * 91.3 s
Calculating the product:
Δθ ≈ 3589.09 rad
Therefore, the angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians.
To convert this angular displacement to revolutions, we can use the conversion factor: 1 revolution = 2π radians.
Δθ in revolutions = Δθ in radians / (2π)
Δθ in revolutions = 3589.09 rad / (2π)
Calculating:
Δθ ≈ 570.46 revolutions
Therefore, the angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians or 570.46 revolutions.
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Light of wavelength 630 nm falls on two slits and produces an Part A interference pattern in which the third-order bright fringe is 40 mm from the central fringe on a screen 3.0 m away. What is the separation of the two slits? Express your answer using two significant figures. X Incorrect; Try Again; 4 attempts remaining
The separation of the two slits is 0.0158 mm (approx).
Given,Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright fringe is 40 mm from the central fringe on a screen 3.0 m away.Concept used,The formula for calculating the separation between the two slits is given by: dsinθ = m λ where, m = 1,2,3,...θ = angle of diffractiond = separation between the slitsλ = wavelength of the lightThe path difference between the waves of light emerging from two slits is given by: d sinθ = mλ where, d = separation between the slitsλ = wavelength of lightθ = angle of diffraction.The third order fringe means m = 3. Hence, d sinθ = 3λHere, λ = 630 nm = 6.3 × 10⁻⁷ m. Therefore,3d sinθ = 6.3 × 10⁻⁷ ...................(1)
We know that angle of diffraction, θ can be given by, θ = tan⁻¹(y/L) where y is the fringe width and L is the distance between the screen and the slits. Here, y = 40 mm = 0.04 mL = 3.0 m Therefore, θ = tan⁻¹ (0.04/3)Now, substitute this value of θ in equation (1), we get:3d × (0.04/3) = 6.3 × 10⁻⁷Or, d = (6.3 × 10⁻⁷ )/0.04The value of d is, d = 1.58 × 10⁻⁵ m or 0.0158 mm (approx).Hence, the separation of the two slits is 0.0158 mm (approx).
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What is the sing of the gravitational potential energy for a 777 jet airliner cruising at 10 000 m above the surface of Earth? The mass of the airliner is 250 000 kg.
The gravitational potential energy of the 777 jet airliner cruising at 10,000 m above the surface of the Earth is 24,500,000,000 joules.
The sign of gravitational potential energy depends on the reference point chosen for the zero potential energy. In the case of the airliner cruising at 10,000 m above the surface of the Earth, we need to consider the conventional choice of the surface of the Earth as the reference point. Gravitational potential energy is defined as the energy possessed by an object due to its position in a gravitational field. It is given by the equation:
PE = mgh,
where PE represents gravitational potential energy, m represents the mass of the object, g represents the acceleration due to gravity, and h represents the height or distance above the reference point.
In this scenario, the airliner is cruising at an altitude of 10,000 m above the surface of the Earth. Since the surface of the Earth is taken as the reference point, the height (h) in the equation will be positive.
m = 250,000 kg (mass of the airliner),
g = 9.8 m/s^2 (acceleration due to gravity),
h = 10,000 m (height above the surface),
PE = 250,000 kg × 9.8 m/s^2 × 10,000 m.
Calculating the value:
PE = 24,500,000,000 J (joules).
Since the height (h) is positive in this case, the gravitational potential energy is also positive. This indicates that the airliner has potential energy relative to the chosen reference point, which in this case is the surface of the Earth.
Therefore, the gravitational potential energy of the 777 jet airliner cruising at 10,000 m above the surface of the Earth is 24,500,000,000 joules.
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