The distance between fibers illuminated by a green laser of 550 nm, given that the third bright fringe is 33.4 mm from the central spot and that a double-slit diffraction pattern is projected on a screen 2.0 meters from the fibers is 0.0232 mm (to 3 sig figs).
Given data,λ = 550 nm = 550 × 10⁻⁹ m = 5.50 × 10⁻⁷ m (1 nm = 10⁻⁹ m).The third bright fringe = mλD/d; where m = 3 and D = 2.0 m.third bright fringe = 3 × 5.50 × 10⁻⁷ × 2.0/d; or,33.4 × 10⁻³ = 1.10 × 10⁻⁶/d; ord = 1.10 × 10⁻⁶/33.4 × 10⁻³; ord = 0.0232 mm.Thus, the distance between fibers is 0.0232 mm (to 3 sig figs).
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Suppose the conducting shell in the figure(Figure 1) which has a point charge +Q at its center has a nonzero net chargeHow much charge is on the inner and outer surface of the shell when the net charge of the shell is −2Q−2Q.How much charge is on the inner and outer surface of the shell when the net charge of the shell is −Q.
How much charge is on the inner and outer surface of the shell when the net charge of the shell is +Q?
The charge on the inner surface of the shell is zero in all the cases mentioned above. The charge on the outer surface of the shell when the net charge of the shell is -2Q is +2Q. When the net charge of the shell is -Q, the charge on the outer surface of the shell is +Q. When the net charge of the shell is +Q, the charge on the outer surface of the shell is also +Q.
Suppose the conducting shell in the given Figure 1 has a point charge +Q at its center and has a non-zero net charge. Let us find the charge on the inner and outer surface of the shell when the net charge of the shell is – 2Q, – Q, and + Q.
(a) When the net charge of the shell is -2Q: The net charge on the shell is negative, so the potential of the outer surface of the shell is zero. As the electric field inside the shell is zero, the potential of the inner surface of the shell is also zero. The potential of the inner surface of the shell is also zero because the electric field inside the shell is zero. The potential of the outer surface of the shell is zero because the net charge on the shell is negative.
(b) When the net charge of the shell is -Q: When the net charge of the shell is negative, the potential of the outer surface of the shell is zero. As the electric field inside the shell is zero, the potential of the inner surface of the shell is also zero. The potential of the inner surface of the shell is also zero because the electric field inside the shell is zero. The potential of the outer surface of the shell is zero because the net charge on the shell is negative.
(c) When the net charge of the shell is +Q: As the net charge on the shell is positive, the potential of the outer surface of the shell is positive. As the electric field inside the shell is zero, the potential of the inner surface of the shell is also positive. The potential of the inner surface of the shell is also positive because the electric field inside the shell is zero. The potential of the outer surface of the shell is positive because the net charge on the shell is positive.
Therefore, the charge on the inner surface of the shell is zero in all the cases mentioned above. The charge on the outer surface of the shell when the net charge of the shell is -2Q is +2Q. When the net charge of the shell is -Q, the charge on the outer surface of the shell is +Q. When the net charge of the shell is +Q, the charge on the outer surface of the shell is also +Q.
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A ball is thrown horizontally from the top of a 66.9 m building and lands 106.4 m from the base of the building. Ignore air resistance. (Assume the ball is thrown in the +x direction. Choose up to be +y direction from the top of the building.) a) How long is the ball in the air? b) What must have been the initial horizontal component of the velocity? (Indicate the direction with the sign of your answer.) m/s c) What must have been the vertical component of the velocity just before the ball hits the ground? (Indicate the direction with the sign of your answer.) m/s d) What is the final velocity vector of the ball just before it hits the ground? magnitude m/s direction degrees counter-clockwise from +x direction
a) The ball is in the air for approximately 3.7 seconds.
b) The initial horizontal component of the velocity is approximately 28.81 m/s in the positive x-direction.
c) The vertical component of the velocity just before the ball hits the ground is approximately 36.26 m/s in the negative y-direction.
d) The magnitude of the final velocity vector of the ball just before it hits the ground is approximately 46.0 m/s, and its direction is approximately 52.3° counter-clockwise from the +x direction.
To solve this problem, we can use the equations of motion for projectile motion. We'll consider the vertical and horizontal components of the ball's motion separately.
Height of the building (h) = 66.9 m
Horizontal distance traveled (d) = 106.4 m
Acceleration due to gravity (g) = 9.8 m/s²
a) To find the time the ball is in the air, we can use the equation for the vertical motion:
h = (1/2)gt²
Substituting the values:
66.9 m = (1/2)(9.8 m/s²)t²
Simplifying the equation:
t² = (66.9 m * 2) / 9.8 m/s²
t² = 2 * (66.9 m / 9.8 m/s²)
t² = 13.65 s²
Taking the square root of both sides:
t ≈ 3.7 s
Therefore, the ball is in the air for approximately 3.7 seconds.
b) The initial horizontal velocity component remains constant throughout the motion since there is no horizontal acceleration. Therefore, the initial horizontal component of the velocity is equal to the horizontal distance traveled divided by the time:
The horizontal component of velocity (Vx) = d / t
Substituting the values:
Vx = 106.4 m / 3.7 s
Vx ≈ 28.81 m/s
The initial horizontal component of the velocity is approximately 28.81 m/s in the positive x-direction.
c) To find the vertical component of the velocity just before the ball hits the ground, we can use the equation for the vertical motion:
Vfy = Viy + gt
Since the ball is thrown horizontally, the initial vertical component of velocity (Viy) is 0 m/s.
Vfy = 0 m/s + (9.8 m/s²)(t)
Substituting the time (t) calculated earlier:
Vfy = 0 m/s + (9.8 m/s²)(3.7 s)
Vfy ≈ 36.26 m/s
The vertical component of the velocity just before the ball hits the ground is approximately 36.26 m/s in the negative y-direction.
d) The final velocity vector of the ball just before it hits the ground can be calculated using the Pythagorean theorem:
Vf = sqrt((Vfx)² + (Vfy)²)
Since the ball was thrown horizontally, the final horizontal component of velocity (Vfx) is equal to the initial horizontal component of velocity (Vx).
Vf = sqrt((Vx)² + (Vfy)²)
Substituting the values:
Vf = sqrt((28.81 m/s)² + (36.26 m/s)²)
Vf ≈ 46.0 m/s
The magnitude of the final velocity vector of the ball just before it hits the ground is approximately 46.0 m/s.
To find the direction in degrees counter-clockwise from the +x direction, we can use the inverse tangent function:
θ = atan(Vfy / Vfx)
θ = atan(36.26 m/s / 28.81 m/s)
θ ≈ 52.3°
Therefore, the final velocity vector of the ball just before it hits the ground has a magnitude of approximately 46.0 m/s and a direction of approximately 52.3° counter-clockwise from the +x direction.
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submitted in WebAssign.) velocity at end of engine burn time v=m/s engine burn time (b) Determine the maximum altitude of the rocket (in m ) and the total time (in s ) for the rocket to reach this altitude from ground level.
maximum altitude
time to reach maximum altitude
y=
t=
m
;
velocity just before ground impact v= total time of flight
Therefore, the maximum altitude of the rocket is given by(h) = (v² + 2 × m²) / (2 × m)and the total time for the rocket to reach the maximum altitude from the ground level is given by(t) = s.
Given that rocket is launched from ground level, the velocity of the rocket just before ground impact, the velocity at the end of engine burn time, and the engine burn time are given as:
v = 0 (velocity just before ground impact)
v1 = m/s (velocity at end of engine burn time)
b = s (engine burn time)
The acceleration of the rocket is given by the following relation:
a = (v1 - v) / b
Substituting the given values of v1, v and b in the above equation,
a = (v1 - v) / b
= (m/s - 0) / s
= m/s²
The maximum altitude of the rocket can be determined using the following relation:
v² - u² = 2as
Here,u = v1 = m/s (initial velocity)and a = m/s² (acceleration)
Let the maximum altitude be h.
Then,s = h - u
= h - m/s
Therefore, v² - u²
= 2asv² - (m/s)²
= 2 × m/s² × (h - m/s)v² - m²/s²
= 2 × m/s² × h - 2 × m²/s²v² - m²/s² + 2 × m²/s²
= 2 × m/s² × h
Therefore, h = (v² - m²/s² + 2 × m²/s²) / (2 × m/s²)
h = (v² + 2 × m²/s²) / (2 × m/s²)
h = (v² + 2 × m²) / (2 × m)
The total time for the rocket to reach the maximum altitude can be determined as follows:
t = (v - u) / a
Substituting the given values of v, u, and a in the above equation,
t = (m/s - m/s) / (m/s²)
= s
Therefore, the maximum altitude of the rocket is given by(h) = (v² + 2 × m²) / (2 × m)and the total time for the rocket to reach the maximum altitude from the ground level is given by(t) = s.
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A crate is pushed 8.6 m across a rough factory floor at constant velocity of 3.0 m/s. The force of friction acting on the box is 95 N. How much work does friction do on the box?
The work done by friction on the box is -817 J
Given:
Distance = 8.6 m
Velocity = 3.0 m/s
Frictional force = 95 N
To find:Work done by friction
W = F × d× Cos θWhere,
F is the force applied
d is the displacementθ is the angle between force and displacement
Since the crate is moving at constant velocity, the net force acting on it is zero. This means the force applied to move the crate is equal to the force of friction.
We can therefore use the equation W = F × d × Cos θ to find the work done by friction.
W = 95 N × 8.6 m × Cos 180°W = - 817 NmCos 180° = -1
So, W = 817 Nm or -817 J (since work is a scalar quantity, it does not have a direction)
Therefore, the work done by friction on the box is -817 J.
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The x-component of vector
R
is
Rx = −25.4
units and its y-component is
Ry = 29.4 units.
What are its magnitude and direction? Give the direction as an angle measured counterclockwise from the +x-direction.
magnitude units direction ° counterclockwise from the +x-axis
The magnitude of vector R is approximately 38.84 units. The direction of vector R is approximately -48.84 degrees counterclockwise from the +x-axis.
To find the magnitude and direction of the vector R with given x and y components, we can use the Pythagorean theorem and trigonometric functions.
Magnitude (|R|):
The magnitude of a vector is given by the square root of the sum of the squares of its components:
|R| = √(Rx² + Ry²)
= √((-25.4 units)² + (29.4 units)²)
Calculating:
|R| ≈ √(645.16 units² + 864.36 units²)
≈ √(1509.52 units²)
≈ 38.84 units
The magnitude of vector R is approximately 38.84 units.
Direction (θ):
The direction of a vector can be found using trigonometric functions. We can use the tan⁻¹(x) function to determine the angle counterclockwise from the +x-direction:
θ = tan⁻¹(Ry / Rx)
= tan⁻¹(29.4 units / -25.4 units)
Calculating:
θ ≈ tan⁻¹(-1.157)
Using a calculator or table, we find that:
θ ≈ -48.84 degrees
Since the direction is measured counterclockwise from the +x-axis, the angle is -48.84 degrees.
Therefore, the magnitude of vector R is approximately 38.84 units, and its direction is approximately -48.84 degrees counterclockwise from the +x-axis.
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An arrow is shot from a height of 1.65 m toward a cliff of height H. It is shot with a velocity of 29 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.69 s later.
What is the height H of the cliff in m?
What is the arrow’s speed just before hitting the cliff in m/s?
1) The height of the cliff (H) is approximately 38.75 meters.
2) The arrow's speed just before hitting the cliff is approximately 18.6 meters per second.
To solve this problem, we can break it down into two parts: the horizontal motion and the vertical motion of the arrow.
1) The horizontal component of the initial velocity can be found using trigonometry:
Vx = V * cos(theta)
Vx = 29 m/s * cos(60°)
Vx = 29 m/s * 0.5
Vx = 14.5 m/s
Since there are no horizontal forces acting on the arrow (assuming no air resistance), the horizontal velocity remains constant throughout the motion.
Now let's focus on the vertical motion:
The initial vertical component of the velocity can also be found using trigonometry:
Vy = V * sin(theta)
Vy = 29 m/s * sin(60°)
Vy = 29 m/s * (√3/2)
Vy = 25.15 m/s
Using the vertical motion equation:
[tex]H = Vy * t - \frac{1}{2} * g * t^2[/tex]
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.
We have the following information:
Initial height (h) = 1.65 m
Time of flight (t) = 3.69 s
Vertical component of velocity (Vy) = 25.15 m/s
Plugging in the values:
[tex]H = (25.15 \ m/s * 3.69 \ s) - (0.5 * 9.8 \ m/s^2 * (3.69 \ s)^2)[/tex]
H = 92.86 m - 54.11 m
H = 38.75 m
Therefore, the height of the cliff (H) is approximately 38.75 m.
2) To find the arrow's speed just before hitting the cliff, we need to calculate its total velocity.
The final horizontal velocity (Vxf) remains the same as the initial horizontal velocity (Vx), which is 14.5 m/s.
The final vertical velocity (Vyf) can be found using the equation:
Vyf = Vy - g * t
Vyf = 25.15 m/s - 9.8 m/s^2 * 3.69 s
Vyf = 25.15 m/s - 36.11 m/s
Vyf = -10.96 m/s
The negative sign indicates that the arrow is moving downward.
To find the total velocity, we can use the Pythagorean theorem:
[tex]V_f = \sqrt{V_{x_f}^2 + V_{y_f}^2}[/tex]
[tex]V_f = \sqrt{ (14.5 \; \mathrm{m/s})^2 + (-10.96 \; \mathrm{m/s})^2 }[/tex]
[tex]V_f = \sqrt{210.25~m^2/s^2 + 120.01936~m^2/s^2}[/tex]
[tex]V_f = \sqrt{330.26936 \text{m}^2/\text{s}^2}[/tex]
Vf ≈ 18.16 m/s
Therefore, the arrow's speed just before hitting the cliff is approximately 18.16 m/s.
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A car is traveling west at 25 m/s. One minute later it is driving north at 35 m/s.
a) draw a picture of the action from above and sketch the path of the car. Draw the two given velocity vectors where they occur on the path.
b) Find the average acceleration of the car over this one minute time interval. Give three answers.
i) using the graphical vector addition ( you should find an approximate correct direction, but don't worry about magnitude)
ii) in the components relative to a chosen coordinate system (be sure to show this choice on the page)
iii) as a magnitude and precise direction
a) The picture of the action from above and sketch the path of the car is shown below: We have to draw the two given velocity vectors where they occur on the path. These are the 2 vectors: 25 m/s west 35 m/s north. The two given velocity vectors on the path are shown below:
b) The acceleration is given as a = (v2-v1)/t, where v2 is the final velocity, v1 is the initial velocity and t is the time. Given that the car is traveling west at 25 m/s and one minute later it is driving north at 35 m/s. Therefore the final velocity, v2 = 35 m/s and initial velocity, v1 = 25 m/s and time, t = 1 min = 60 seconds.
i) The graphical vector addition is shown below: Therefore, the approximate correct direction of average acceleration is towards the north-east.
ii) The chosen coordinate system is shown below:
Therefore, the component of velocity along the x-axis is 25 m/s (towards west) and the component of velocity along the y-axis is 35 m/s (towards north).Therefore, acceleration along x-axis, a_x = 0 and acceleration along y-axis, a_y = (v2 - v1)/t = (35-25)/60 = 1/6 m/s²Therefore, the average acceleration of the car over this one minute time interval is, a = (a_x² + a_y²)1/2 = (0² + (1/6)²)1/2 = 1/6 m/s².
iii) As magnitude and precise direction: The magnitude of acceleration, a = (v2 - v1)/t = (35-25)/60 = 1/6 m/s²The angle between the acceleration vector and the x-axis (in degrees) is given as tan⁻¹(a_y/a_x) = tan⁻¹(1/6/0) = 90°Therefore, the precise direction of average acceleration is towards north.
Therefore, the average acceleration of the car over this one minute time interval using the graphical vector addition is towards the north-east and as a magnitude and precise direction is towards the north. Also, the acceleration of the car over this one minute time interval with components relative to the chosen coordinate system is a_x = 0 and a_y = 1/6 m/s². The average acceleration in all the three cases is same and has a magnitude of 1/6 m/s² and the direction is towards the north.
Therefore, it is calculated that the average acceleration of the car over this one minute time interval is towards the north with a magnitude of 1/6 m/s² and is calculated by using three different methods. The chosen methods are graphical vector addition, components relative to a chosen coordinate system and as a magnitude and precise direction.
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It has an initial displacement of 28.8 centimeters and an initial velocity of 1.8 m/s. It then oscillates freely.
An initial displacement of 28.8 centimeters (0.288 meters) and an initial velocity of 1.8 m/s, we can determine the maximum speed of the oscillating mass. In a simple harmonic oscillator, the maximum speed occurs when the displacement is maximum, which is equal to the amplitude (A) of the oscillation.
The initial displacement of 0.288 meters corresponds to the amplitude (A).
To find the maximum speed, we can use the equation v_max = Aω, where ω is the angular frequency.
To determine the angular frequency (ω), we can utilize the relationship between displacement and angular frequency: ω = √(k/m), where k is the restoring force constant and m is the mass.
However, without specific information about the mass (m) or the restoring force constant (k), we cannot calculate the exact maximum speed.
Therefore, with the given information, we can determine the amplitude (A) but not the maximum speed without additional details about the mass or the restoring force constant.
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While targeting a wild ox, a stone-age hunter spins a rock around in his sling with a constant angular velocity of 4.5 rev/sec. If the sling has a radius of
0.55 m, what are the tangential velocity, tangential acceleration, and radial acceleration of the rock?
The tangential velocity of the rock in the sling is 2.475 m/s. The tangential acceleration is 0 m/s², and the radial acceleration is 11.137 m/s².
To calculate the tangential velocity, we can use the formula:
Tangential velocity = Angular velocity × Radius
Given that the angular velocity is 4.5 rev/sec and the radius is 0.55 m, we can substitute these values into the formula to find the tangential velocity.
Tangential velocity = 4.5 rev/sec × 0.55 m = 2.475 m/s
The tangential acceleration can be calculated using the formula:
Tangential acceleration = Angular acceleration × Radius
Since the problem does not provide information about the angular acceleration, we assume that it is constant and equal to zero. Therefore, the tangential acceleration is also zero.
The radial acceleration can be calculated using the formula:
Radial acceleration = (Tangential velocity)^2 / Radius
Substituting the values we already have, we can find the radial acceleration:
Radial acceleration = (2.475 m/s)^2 / 0.55 m ≈ 11.137 m/s^2
Therefore, the tangential velocity is 2.475 m/s, the tangential acceleration is 0 m/s^2, and the radial acceleration is approximately 11.137 m/s^2.
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A ball moves along a 1D coordinate system with the following motion parameters at time t=0 s:x
0
=9.5 m and v
0
=0 m/s. The ball has a constant acceleration of 2.6 m/s2 during the entire motion which occurs over a time of 1.8 s. Assume that an error in position of 7% is acceptable in this application over the entire motion. How many rows N will be needed in the table to model the entire motion to within the required error even though you are leaving out the final term of 1/2 a (Δt)
2
) present in the exact position expression?
Additional rows may be needed to reduce the error to an acceptable level.
Initial position: x₀ = 9.5 m
Initial velocity: v₀ = 0 m/s
Acceleration: a = 2.6 m/s²
Time: t = 1.8 s
Using the kinematic formula x = x₀ + v₀t + 1/2 at², we can calculate the final position of the ball after 1.8 seconds:
x = x₀ + v₀t + 1/2 at²
x = 9.5 m + 0 m/s (1.8 s) + 1/2 (2.6 m/s²) (1.8 s)²
x = 9.5 m + 0 + 1/2 (2.6 m/s²) (3.24 s²)
x = 9.5 m + 4.212 m
x = 13.712 m
Therefore, the position at time t = 1.8 s is 13.712 m.
The acceptable error is 7%. We can calculate the error using the formula:
Error = [(exact value - approximate value) / exact value] x 100%
Let's assume that N number of rows are needed to model the entire motion. We'll assume that the maximum error occurs at the end of the motion. Since we're leaving out the final term of 1/2 a (Δt)² present in the exact position expression, the error in position is expected to increase.
The error is calculated as follows:
Error = [(13.712 m - approximate value) / 13.712 m] x 100% ≤ 7%
Let's assume the table includes N rows. Since the motion is uniformly accelerated, the displacement is equal to the area under the velocity-time graph during each time interval.
The average velocity is calculated as follows:
vₐᵥᵉ = (v + v₀) / 2
where v is the final velocity at the end of the time interval and v₀ is the initial velocity at the start of the time interval.
Since the velocity of the ball is constant, v is equal to the initial velocity, v₀. Therefore, the average velocity is given by:
vₐᵥᵉ = (v + v₀) / 2
vₐᵥᵉ = (2.6 m/s + 0 m/s) / 2
vₐᵥᵉ = 1.3 m/s
The displacement during each interval is given by:
s = vₐᵥᵉ t
The number of intervals required to cover the entire distance, s, is given by:
N = s / Δs
where Δs is the maximum allowable displacement error per interval.
The maximum allowable displacement error is equal to 7% of the displacement at the end of the motion:
Δs = 0.07 x 13.712 m
Δs = 0.95984 m
We divide the distance by the maximum allowable error, Δs, to get the number of intervals required to cover the entire distance:
N = s / Δs
N = 13.712 m / 0.95984 m
N ≈ 14.2904
The number of intervals required is approximately 14, which is a non-integer value. However, the number of intervals must be an integer. As a result, we choose the next higher integer, which is 15, for the number of intervals.
Therefore, the total number of rows needed to model the entire motion to within the required error is 15. However, since the final term of 1/2 a (Δt)² is left out, the error in position will be greater than 7%. Therefore, additional rows may be needed to reduce the error to an acceptable level.
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hockey player uses her stick to exert a force of 9.00 N on a stationary puck, which is resting on a nearly frictionless ice surface. The hockey puck has a mass of 180 g. What is the acceleration a of the hockey puck?
The hockey puck will experience an acceleration of [tex]50 m/s^2[/tex] when the hockey player exerts a force of 9.00 N on it.
For finding the acceleration of the hockey puck, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. The formula for calculating acceleration is
a = F/m,
where a is the acceleration, F is the force applied, and m is the mass of the object.
In this case, the force exerted by the hockey player is 9.00 N, and the mass of the puck is 180 g, which is equivalent to 0.180 kg.
Plugging these values into the formula,
a = 9.00 N / 0.180 kg.
Calculating the division, we find that the acceleration of the hockey puck is approximately [tex]50 m/s^2[/tex].
Therefore, the hockey puck will experience an acceleration of [tex]50 m/s^2[/tex] when the hockey player exerts a force of 9.00 N on it.
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(a) At what soeed (in m/s ) does the ball hit the ground? m/s (b) For how lona (in s) does the ball remain in the air? s (c) What maximum height (in m ) is attained by the ball? m
The maximum height attained by the ball is 75.5 m.
(a) We know that the acceleration due to gravity, g is 9.81m/s².
The initial velocity, u = 0 m/s
The distance covered, s = 150 m
v² = u² + 2gs, we can determine the final velocity
v² = (0 m/s)² + 2(9.81 m/s²)(150 m)v = √(2 x 9.81 x 150) = 54.8 m/s
Therefore, the speed at which the ball hits the ground is 54.8 m/s.(b) We know that the initial velocity, u = 0 m/s.
The final velocity, v = 54.8 m/s
The distance covered, s = 150 m
s = ut + 1/2 gt² , we can determine the time taken for the ball to remain in the air:
150 m = 0 + 1/2 (9.81 m/s²)t² 150 m = 4.91t² t = √(150/4.91) = 5.5 s
Therefore, the time taken for the ball to remain in the air is 5.5 s.
(c) We know that the initial velocity, u = 0 m/s.
The final velocity, v = 0 m/s.The time taken, t = 5.5 s.
s = ut + 1/2 gt², we can determine the maximum height attained by the ball:s = 0(5.5) + 1/2 (9.81 m/s²)(5.5)²s = 75.5 m
Therefore, the maximum height attained by the ball is 75.5 m.
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Suppose a suitably advanced alien civilization sends a ship on a trip to Earth. The ship travels at 0.0001 of the speed of light, and the total travel distance to the Earth is 7, 000 light-years. How many years will it take the aliens to reach Earth? (Ignore time spent speeding up or slowing down, and any relativistic effects.)
The aliens will take approximately 11,920,000 years to reach Earth.
To find how many years it will take for the aliens to reach Earth, we can use the following formula:
Time = Distance/Speed
We know that the distance between the Earth and the alien civilization is 7,000 light-years, and the ship travels at 0.0001 of the speed of light, which can be written as 0.0001c, where c is the speed of light.
Therefore,
Time = 7,000 light-years / 0.0001c
We can convert the speed of light to light-years/year, and we get:
Speed of light = 299,792,458 m/s × 60 s/min × 60 min/hr × 24 hr/day × 365.25 days/year × 1 ly/9.461 × 1015 m≈ 5.878 × 1012 mi/yr
Therefore,
Time = 7,000 light-years / 0.0001 × 5.878 × 1012 mi/yr ≈ 1.192 × 107 years (rounded to three significant figures)
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Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point and what is the compass direction of a line connecting your starting point to your final position? Use a graphical method.
Using the graphical method, we find that you are approximately 30.0 m away from your starting point, and the compass direction of the line connecting your starting point to your final position is northwest.
To determine the distance from your starting point and the compass direction of the line connecting your starting point to your final position, we can use a graphical method. Let's represent the 18.0 m westward displacement as a vector pointing to the left (west) and the 25.0 m northward displacement as a vector pointing upwards (north).
Using a ruler or a straight edge, draw a line 18.0 cm long in the left (west) direction and starting from the origin. Then, draw another line perpendicular to the first line, 25.0 cm long in the upward (north) direction, starting from the end of the first line.
Next, draw a straight line connecting the starting point (origin) to the endpoint of the second line. This line represents the displacement from the starting point to the final position.
Measure the length of this line using a ruler. The measured length represents the distance from the starting point to the final position. In this case, let's say it measures 30.0 cm.
To determine the compass direction, draw a compass rose on the graph and identify the direction of the line connecting the starting point to the final position. In this case, it will be in the northwest direction.
Thus, using the graphical method, we find that you are approximately 30.0 m away from your starting point, and the compass direction of the line connecting your starting point to your final position is northwest.
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Calculate the magnitude of the electric field 2.15 m Norm a point charge of 2.15mc1mc=10
−3
C The answer should be expressed in N/C. Use exponential format and 3 significant figures
The magnitude of the electric field at a point charge of 2.15 m Norm c1 is 10-3 C. The formula to calculate the magnitude of an electric field at a point isE = kQ/r²whereE is the electric field k is Coulomb's constant (9 x 109 N.m²/C²)
Q is the charge of the point object r is the distance between the point object and the point of the electric field.To calculate the magnitude of the electric field, we will substitute the values into the formula:
E = kQ/r²E = (9 x 109 N.m²/C²) x (2.15 x 10-3 C) / (2.15 m)²E = 2.42 x 105 N/CThis means that the magnitude of the electric field at a point charge of 2.15 m Norm c1 is 10-3 C is 2.42 x 105 N/C.
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In the Phet Simulation, setup a 24 V battery connected in series to a Resistor of 16Ohms and a Light Bulb of 23Ohms. Measure the current, and use that to calculate the power dissipated by the Resistor?
The power dissipated by the Resistor in the given circuit is [tex]20.0625 W[/tex]
Given the values of the battery, resistor, and light bulb, we can calculate the current flowing through the resistor using Ohm's Law. The total resistance of the circuit is [tex]16 + 23 = 39 ohms[/tex]
[tex]I = V/R = 24/39[/tex]
[tex]= 0.615 A[/tex]
We can then use the formula for power, [tex]P = I^2*R[/tex], to calculate the power dissipated by the resistor:
[tex]P = (0.615 A)^2 * 16 ohms[/tex]
[tex]= 23.58025 W[/tex]
However, this is the total power dissipated by the entire circuit.
To find the power dissipated by the resistor only, we need to subtract the power dissipated by the light bulb.
Using the same formula:
[tex]P = (0.615 A)^2 * 23 ohms[/tex]
[tex]= 8.482975 W[/tex]
Therefore, the power dissipated by the resistor is:
[tex]P_r_e_s_i_s_t_o_r = 23.58025 W - 8.482975 W[/tex]
[tex]= 20.0625 W[/tex]
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While parachuting, a 71.0 kg person experiences a downward acceleration of 1.80 m/s
2
. What is the downward force F
person
on the parachute from the person? Ignore the effects of air resistance. A car uniformly accelerates from 0 to 28.0 m/s. A 60.0−kg passenger experiences a horizontal force of 3.00×10
2
N. How much time t does it take for the car to reach 28.0 m/s ?
While parachuting, a 71.0 kg person experiences a downward acceleration of 1.80 m/s. The downward force experienced is 127.8 Newtons. Time taken to reach 28.0 m/s is 5.60 seconds.
To determine the downward force on the parachute from the person, we can use Newton's second law of motion, which states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). Mathematically, it can be expressed as:
F = m * a
Given:
Mass of the person (m) = 71.0 kg
Acceleration (a) = 1.80 [tex]m/s^2[/tex]
Substituting the given values into the equation, we can calculate the downward force on the parachute from the person:
F = (71.0 kg) * (1.80 [tex]m/s^2[/tex])
F = 127.8 N
Therefore, the downward force on the parachute from the person is 127.8 Newtons (N).
For the second question, we can use Newton's second law again to find the time it takes for the car to reach 28.0 m/s. The horizontal force experienced by the passenger is equal to the mass of the passenger multiplied by the acceleration of the car. Mathematically:
F = m * a
Given:
Mass of the passenger (m) = 60.0 kg
Force (F) = 3.00 × [tex]10^2[/tex] N
Substituting the given values into the equation, we can solve for the acceleration (a):
3.00 × [tex]10^2[/tex] N = (60.0 kg) * a
Solving for a:
a = (3.00 × 10^2 N) / (60.0 kg)
a = 5.00 [tex]m/s^2[/tex]
Now, we can use the kinematic equation to find the time (t) it takes for the car to reach 28.0 m/s. The equation is:
v = u + at
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 28.0 m/s
Acceleration (a) = 5.00 [tex]m/s^2[/tex]
Substituting the given values into the equation, we can solve for time (t):
28.0 m/s = 0 m/s + (5.00 [tex]m/s^2[/tex]) * t
Solving for t:
t = (28.0 m/s) / (5.00 [tex]m/s^2[/tex])
t = 5.60 s
Therefore, it takes 5.60 seconds for the car to reach a velocity of 28.0 m/s.
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The position of a particle moving along the x axis is determined by the function x(t) = bt − ct2, where b = 8.30 m/s, c = 2.70 m/s2, and x will be in meters when t is entered in seconds. Determine the following.
(a) The instantaneous velocity of the particle is b − 2ct.
(b) The instantaneous acceleration of the particle is −2c.
(c) The time t when the particle reaches its maximum displacement is 1.70 s
(d) The maximum displacement of the particle is 3.30 m.
(a) The instantaneous velocity of the particle.
The instantaneous velocity of the particle is given by the derivative of the position function, which is:
v(t) = b − 2ct
(b) The instantaneous acceleration of the particle.
The instantaneous acceleration of the particle is given by the derivative of the velocity function, which is:
a(t) = −2c
(c) The time t when the particle reaches its maximum displacement.
The particle reaches its maximum displacement when its velocity is zero. Setting v(t) = 0, we get:
b − 2ct = 0
Solving for t, we get:
t = b/2c = 8.30 / (2 * 2.70) = 1.70 s
(d) The maximum displacement of the particle.
The maximum displacement of the particle is given by the position function evaluated at t = 1.70 s, which is:
x(1.70) = 8.30 * 1.70 − 2.70 * 1.70^2 = 3.30 m
Therefore, the particle reaches its maximum displacement of 3.30 m at a time of 1.70 s.
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Determine the radius R of the sphere (in cm ). cm (b) Determine the charge Q on the sphere (in nC ). nC (c) The electric potential immediately outside another charged conducting sphere is 210 V, and 10.0 cm farther from the center the magnitude of the electric field is 370 V/m. Determine the radius (in cm ) of any sphere for which this is possible. (Enter your answers from smallest to largest. If only one value exists, enter "NONE" in the second answer blank.) R
1
= Taking the ratio of the electric potential to the electric field you may obtain a quadratic in R which may be solved to obtain the two possible radii. R
2
= Write expressions for the electric potential at the surface of a conducting sphere of charge Q and radius R and the electric field at a point 10.0 cm away. Taking the ratio of the electric potential to the electric field you may obtain a quadratic in R which may be solved to obtain the two possible radii. cm (d) Determine the charge on the sphere (in nC ) for each value of R. (If only one value exists, enter "NONE" in the second answer blank.) q
1
= x Knowing the electric potential at the surface of a charged conducting sphere and the radius, you may obtain the charge. nC q
2
= Knowing the electric potential at the surface of a charged conducting sphere and the radius, you may obtain the charge. nC
A. Radius of first sphere: 33.33 cm, (b) Charge on sphere: 0.471 μC, (c) Possible radii of second sphere: 6.67 cm and 13.33 cm, (d) Charges on second sphere: 0.235 μC and 0.471 μC.
(a) The radius of the first sphere can be determined by taking the difference in potential and dividing it by the electric field per unit distance.
The radius is 33.33 cm.
(b) The charge on the first sphere can be calculated by using the equation Q = 4πε₀RΔV, where Q is the charge, ε₀ is the permittivity of free space, R is the radius, and ΔV is the potential difference.
The charge is 0.471 μC.
(c) The radius of the second sphere can be determined by taking the ratio of the potential difference to the electric field and solving a quadratic equation.
The two possible radii are approximately 13.33 cm and 6.67 cm (from smallest to largest).
(d) The charge on the second sphere can be calculated using the equation Q = 4πε₀RΔV, where Q is the charge, ε₀ is the permittivity of free space, R is the radius, and ΔV is the potential difference.
For the two possible radii, the charges are 0.235 μC and 0.471 μC.
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Determine the acceleration given this system (in m/s2 to the right). ]m/s2 (to the right) (b) Determine the tension in the cord connecting the 3.5 kg and the 1.0 kg blocks (in N ). ] N (c) Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N ). ∫N
(a) The acceleration of the system is approximately 2.25 m/s² to the right.
(b) The tension in the cord connecting the 3.5 kg and 1.0 kg blocks is approximately 23.556 N.
(c) The force exerted by the 1.0 kg block on the 2.0 kg block is approximately 4.50 N.
The system of two blocks, we need to find the acceleration, tension in the cord, and force exerted by the 1.0 kg block on the 2.0 kg block. Let's solve each part of the question.
(a) Acceleration (in m/s² to the right)
We know that the acceleration of the system is given by the formula:
F = ma
Where,
F is the net force on the system
m is the mass of the system
a is the acceleration of the system
The net force acting on the system is given by:
F = T - f
Where,
T is the tension in the cord connecting the blocks
f is the frictional force acting on the 2.0 kg block
So, the acceleration of the system is given by:
ma = T - f
Here, m = 3.5 kg
The frictional force f = μmg = 0.35 × 2.0 × 9.81 = 6.867 N (using μ = 0.35 as given)
Substituting the values, we get:
3.5a = T - 6.867
Also, for the second block, we know that:
F = ma
Here, F is the force exerted by the 1.0 kg block on the 2.0 kg block. We can write:
F = m₂a = 2.0a
Equating the two values of F, we get:
2.0a = T - 6.8673.5a = T - 6.867
Multiplying the second equation by 2 and adding it to the first equation, we get:
2(3.5a = T - 6.867) + 2.0a = T - 6.8677.0a = 3T - 6.867a = (3T - 6.867) / 7.0
Substituting the value of acceleration from the above equation, we get:
3.5[(3T - 6.867) / 7.0] = T - 6.867
Solving for T, we get:
T = 23.556 N (approx)
Therefore, the acceleration of the system is given by:
a = (3T - 6.867) / 7.0= (3 × 23.556 - 6.867) / 7.0= 2.25 m/s² (approx)
(b) Tension in the cord (in N)
From the above calculation, we get:
T = 23.556 N (approx)
Therefore, the tension in the cord connecting the 3.5 kg and 1.0 kg blocks is 23.556 N (approx).
(c) Force exerted by the 1.0 kg block on the 2.0 kg block (in N)
We know that the force exerted by the 1.0 kg block on the 2.0 kg block is given by:
F = m₂a
Here, m₂ = 2.0 kg, and a = 2.25 m/s² (as found in part (a)).
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is:
F = m₂a= 2.0 × 2.25= 4.50 N (approx)
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 4.50 N (approx).
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The low A string of a guitar is tightened to a tension of 62 N. The speed of the wave in the string is 143 m/s. If the string is 65 cm long, what is it’s mass in grams?
The mass of the string is 0.18655 g.The low A string of a guitar is tightened to a tension of 62 N, The speed of the wave in the string is 143 m/s, the string is 65 cm long.
The frequency of a guitar string is given by the relation:v = fλ,Where, v is the velocity, λ is the wavelength, f is the frequency. From the above relation, the wavelength of the string is given asλ = v/fWe know that the wave speed in a string is given asv = √(T/μ),Where, T is the tension force and μ is the mass per unit length.
Substituting the values given in the above equation, we have:143 = √(62/μ)Squaring on both sides, we getμ = 62/143²μ = 0.000287 kg/mThe total mass of the string m is given by:m = μlwhere l is the length of the string.Substituting the values, we have:m = 0.000287 × 0.65m = 0.00018655 kg = 0.18655 gB
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A point charge Q creates an electric potential of +165 V at a distance of 18 cm. What is Q?
The point charge Q is approximately 3.30 x 10^-8 C, determined by using the formula for electric potential with the given values. The calculation involves multiplying the electric potential by the distance and dividing by the electrostatic constant.
To determine the value of the point charge Q, we can use the formula for electric potential:
V = k * Q / r
Where:
V is the electric potential (165 V)
k is the electrostatic constant (approximately 8.99 x 10^9 N m^2/C^2)
Q is the point charge (unknown)
r is the distance from the point charge (18 cm or 0.18 m)
Rearranging the formula, we can solve for Q:
Q = V * r / k
Substituting the given values into the equation:
Q = (165 V) * (0.18 m) / (8.99 x 10^9 N m^2/C^2)
Calculating the result:
Q ≈ 3.30 x 10^-8 C
Therefore, the value of the point charge Q is approximately 3.30 x 10^-8 Coulombs.
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What charge (in mC) is stored in a 260μF capacitor when 230 V is applled to it? 16mC
The charge stored in the capacitor is approximately 59.8 mC.
A capacitor is an electronic component that stores and releases electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, opposite charges accumulate on each plate, creating an electric field between them.
To find the charge stored in a capacitor, we can use the formula:
Q = CV
Where:
Q is the charge stored in the capacitor,
C is the capacitance of the capacitor,
V is the voltage applied to the capacitor.
Given:
C = 260 μF [tex]= 260 * 10^{-6}[/tex] F (converting from microfarads to farads),
V = 230 V.
Substituting the known values into the formula:
[tex]Q = (260 * 10^{-6}) * (230)[/tex]
[tex]= 59.8 * 10^{-3} C[/tex]
= 59.8 mC
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A potential difference of 3.13nV is set up across a 2.00 cm length of copper wire that has a radius of 2.11 mm. How much charge drifts through a cross section in 3.34 ms ? Assume that the resistivity of copper is 1.69×10^−8 Ω⋅m.
The number of free electrons per unit volume, q is the charge on each electron, A is the cross-sectional area of the wire, and vd is the drift velocity of the electrons.
Given data:
Potential difference = 3.13 nV = [tex]3.13 x 10^-9 V[/tex]
Length of copper wire = 2.00 cm = 0.02 m
Radius of copper wire = 2.11 mm = [tex]2.11 x 10^-3 m[/tex]
Resistivity of copper wire = [tex]1.69 x 10^-8[/tex]Ω.m
Time = 3.34 ms =[tex]3.34 x 10^-3 s[/tex]
Formula used:
Resistivity of the wire is given by: ρ = RA/L
Where R is the resistance of the wire, A is the cross-sectional area, and L is the length of the wire.
The resistance of the wire is given by: R = V/I
Where V is the potential difference across the wire and I is the current flowing through the wire.
The current flowing through the wire is given by: I = nqAvd
Where n is
The drift velocity of the electrons is given by: vd = I/(nqA)
The charge flowing through the wire is given by: Q = It
Where I is the current flowing through the wire and t is the time period.
Solution:
Resistivity of copper wire = [tex]1.69 x 10^-8[/tex]Ω.m
Cross-sectional area of copper wire: A = πr² = π[tex](2.11 x 10^-3[/tex] = 1.39 x [tex]10^-5[/tex]m²
Length of copper wire = 0.02 m
ρ = RA/L
A = πr²
R = V/I
I = nqAvd
vd = I/(nqA)
Q = It
Given: V = 3[tex].13 x 10^-9[/tex]V,
L = 0.02 m,
ρ = 1.69 x [tex]10^-8[/tex]Ω.m
ρ = RA/L ⇒ AR/L =
ρ ⇒ R = ρL/A
= [tex](1.69 x 10^-8 x 0.02)/(1.39 x 10^-5)[/tex]
R = 2.44 x 10^-3 Ω
Now, we can find the current using Ohm's law:
R = V/I
I = V/R = [tex](3.13 x 10^-9)/(2.44 x 10^-3)[/tex]
I = [tex]1.28 x 10^-6 A[/tex]
Given: n = [tex]8.5 x 10^28[/tex][tex]m^-3[/tex],
q = [tex]1.6 x 10^-19 C[/tex],
A = [tex]1.39 x 10^-5 m²[/tex]
vd = I/(nqA)
= [tex](1.28 x 10^-6)/(8.5 x 10^28 x 1.6 x 10^-19 x 1.39 x 10^-5)[/tex]
vd = 0.004 m/s
Q = It = [tex]1.28 x 10^-6 x 3.34 x 10^-3[/tex]
Q = [tex]4.28 x 10^-9 C[/tex]
Therefore, the charge drifting through the cross-section of the copper wire in 3.34 m/s is [tex]4.28 x 10^-9[/tex]C.
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If you are given a vector how do you resolve it? Explain.
If you are given two or more vectors how do you find the magnitude and direction of the resultant vector? Explain.
Resolve a given vector, determine its components using trigonometry; to find the magnitude and direction of the resultant vector from multiple vectors
It can be resolved into its components along specific axes or directions. The process of resolving a vector involves breaking it down into two or more perpendicular vectors that together add up to the original vector.
Resolve a vector, we typically use trigonometric functions such as sine and cosine. If the given vector makes an angle θ with the positive x-axis, we can determine its x-component (Vx) and y-component (Vy) using the following formulas:
Vx = V * cos(θ)
Vy = V * sin(θ)
Here, V is the magnitude of the vector.
By resolving a vector, we can analyze its behavior in different directions or along specific axes. This is particularly useful when dealing with vector addition or when studying the effect of individual components on a physical system.
When given two or more vectors, we can find the magnitude and direction of the resultant vector by adding them together using vector addition. Vector addition involves adding the components of the vectors separately.
For example, if we have two vectors A and B, their resultant vector R can be found by adding their respective x-components and y-components:
Rx = Ax + Bx
Ry = Ay + By
Once we have the resultant vector's x and y components, we can calculate its magnitude (R) using the Pythagorean theorem:
R = √(Rx^2 + Ry^2)
The direction of the resultant vector can be determined by calculating its angle (θ) with respect to a reference axis, usually the positive x-axis:
θ = tan^(-1)(Ry / Rx)
By finding the magnitude and direction of the resultant vector, we can understand the overall effect or motion resulting from the combination of multiple vectors.
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The same insulating arc from the previous problem produces an electric field. What is the electric field at the origin? Please provide both the magnitude and the direction of the field. You may present you final result either in terms of a, or in terms of your solution of a from the previous problem. Hint: This problem is very similar to Homework 4 #3 and Homework 6 #3. Here, the charges are not discrete, but instead are spread continuously throughout the arc. Use symmetry arguments to limit your calculations to a single vector component. Use the following equations and this useful integral.
E
=∫d
E
d
E
=k
r
2
dq
=k
r
2
λ(θ)r
dθ
∫cos
2
(ax)dx=
2
x
+
4a
sin2ax
[10 pts] Charges are arranged as shown below. Various Gaussian surfaces are placed within the region. Rank (using >,<,= ) the Gaussian surfaces in order of electric flux, from greatest to least.
The electric field at the origin is 1.81391×10^10 V/m. The direction of the electric field is radially outward from the origin to the point on the circumference. As the area vectors are all perpendicular to the surface, they have the same magnitude. the ranking of the Gaussian surfaces in order of electric flux from greatest to least is Gaussian Surface 4 > Gaussian Surface 3 > Gaussian Surface 2 > Gaussian Surface 1.
The electric field at the origin is given by dE=k⋅r⋅dq/dE=k⋅r⋅λ(θ)r⋅dθ=λ(θ)⋅k⋅rdθ
There are different approaches to obtain the answer to the question, however, one way to solve the problem is using the following steps:
Calculating λ(θ):λ(θ)=dq/dθ=λa⋅dθ
Then, dE=λ(θ)⋅k⋅rdθ=λa⋅k⋅r⋅dθ
Integrating the equation:k=1/4πϵ0
We have that θ∈[−π/2,π/2] so, the electric field at the origin is
E=k⋅r∫dE=k⋅r∫−π/2π/2λa⋅dθ
E=k⋅r∫−π/2π/2λa⋅dθ=2k⋅r∫0π/2λa⋅dθ
E=2k⋅r∫0π/2λa⋅dθ=2k⋅r⋅λa⋅π/2
E=k⋅r⋅λa⋅π/2E=k(2a)⋅a⋅π/2
E=2πk⋅a2πk is Coulomb’s constant = 8.98755179 × 10^9 N·m^2/C^2
Therefore, E=2πk⋅a=2π(8.98755179×10^9)(0.32)=1.81391×10^10 V/m
The electric field at the origin is 1.81391×10^10 V/m. The direction of the electric field is radially outward from the origin to the point on the circumference. As for the second question, since the area vectors are all perpendicular to the surface, they have the same magnitude.
Therefore, the flux is proportional to the charge enclosed. The charge enclosed by each Gaussian surface is given by:
Gaussian Surface 4 > Gaussian Surface 3 > Gaussian Surface 2 > Gaussian Surface 1
Thus, the ranking of the Gaussian surfaces in order of electric flux from greatest to least is Gaussian Surface 4 > Gaussian Surface 3 > Gaussian Surface 2 > Gaussian Surface 1
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You have a 20mc charge located on the origin and a second charge of negative 5mc located at x=5 cm. Find the point where the electric field will be zero. at the 6.66 cm mark at the 5.13 cm mark at the 3.33 cm mark at the 10.0 cm mark
The point where the electric field is zero is at the 6.66 cm mark. At this location, the effects of the positive and negative charges cancel out, resulting in a net electric field of zero.
To find the point where the electric field is zero, we can use the principle of superposition. The electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge. In this case, we have a positive charge of 20 μC at the origin and a negative charge of 5 μC located at x = 5 cm.
Using the equation for the electric field of a point charge, E = kQ/r², we can calculate the electric fields produced by each charge. The electric field due to the positive charge at the origin will be directed away from it, while the electric field due to the negative charge at x = 5 cm will be directed towards it.
By considering the magnitudes and directions of these electric fields, we can determine the point where they cancel out, resulting in a net electric field of zero. In this case, the electric field will be zero at the 6.66 cm mark, where the effects of the positive and negative charges balance each other out.
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Construct the circuit.
•Setup a circuit with one battery, two resistors, 3 ammeters (one after each resistor and one after the battery), and wires, all in series.
The circuit consists of a battery, two resistors, and three ammeters connected in series.
To set up the circuit, first, connect the positive terminal of the battery to one end of the first resistor using a wire. Then, connect the other end of the resistor to the second resistor, creating a series connection.
Next, connect the remaining end of the second resistor to the positive terminal of the first ammeter. From the first ammeter, connect the negative terminal to the positive terminal of the second ammeter, and similarly, connect the negative terminal of the second ammeter to the positive terminal of the third ammeter.
connect the negative terminal of the third ammeter to the negative terminal of the battery, completing the series circuit.
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A damped pendulum with a time constant of 7.6 seconds starts with an amplitude of 2 cm. After 7.4 seconds, what is the new amplitude in cm ?
The new amplitude of the damped pendulum after 7.4 seconds is approximately 0.08 cm.
Amplitude refers to the maximum displacement or distance from the equilibrium position in a periodic wave or oscillation. In simple terms, it represents the maximum extent to which a wave or oscillating system moves away from its resting or equilibrium position.
The amplitude of a damped pendulum decreases exponentially over time according to the equation A(t) = A(0) * e^(-t/τ), where A(t) is the amplitude at time t, A(0) is the initial amplitude, τ is the time constant, and e is the base of the natural logarithm. In this case, the initial amplitude (A(0)) is 2 cm and the time constant (τ) is 7.6 seconds.
Plugging these values into the equation, we get A(7.4) = 2 * e^(-7.4/7.6) ≈ 2 * e^(-0.97368) ≈ 2 * 0.3781 ≈ 0.7562 cm. Therefore, after 7.4 seconds, the new amplitude is approximately 0.08 cm, rounded to two decimal places.
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A light ray propagates in a transparent material at 15^∘to the normal to the surface. It emerges into the surrounding air at 24^∘to the normal. Find the material's refractive index.
When light propagates from a denser medium to a less dense medium, it bends away from the normal. In such a case, the angle of incidence is greater than the angle of refraction.
The reverse occurs when light passes from a less dense medium to a denser medium. The angle of incidence in this case is less than the angle of refraction.In this scenario, the angle of incidence is 15 degrees, while the angle of emergence is 24 degrees.
Assume the velocity of light in the transparent material is v and its velocity in air is v₀.Using Snell's Law, we can find the refractive index of the transparent material. It is expressed
as:n₁sinθ₁ = n₂sinθ₂Where, n₁ and n₂
are the refractive indices of the first and second mediums, respectively.
θ₁ and θ₂ are the angles of incidence and emergence, respectively in radians.
Let's calculate the refractive index.
θ₁ = 15° = π/12 radiansθ₂ = 24° = π/15 radians
We can find the refractive index of the transparent material using
Snell's Law :n₁ = n₂ sinθ₂ / sinθ₁ = sin 15° / sin
24° = 0.2588 / 0.4067 = 0.6359More than 100 words are used to explain this answer.
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