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xenn [34]
1 year ago
15

A technician is working on an MRI machine. To test it, the technician turns on the MRI machine that produces a strong magnetic f

ield, but then he accidentally drops a screwdriver in this magnetic field. As the screwdriver falls, an emf is induced across the length of the metal shaft, L, of the screwdriver. At one instant, the screwdriver falls with a speed of 2.1 m/s and is oriented as shown in figure so that the induced emf is maximized. The magnitude of the magnetic field at this instant is 2.5 T, and the length of the metal shaft, L, is 10 cm. What is the induced emf across the length of the metal shank at this instant?
Physics
1 answer:
AnnZ [28]1 year ago
4 0

Answer:

0.525 V

Explanation:

Noting the given data in the question:

we have,

The speed of the fall of the screw driver, v = 2.1 m/s

The magnitude of the magnetic field, B = 2.5 T

The length of the metal shaft of the screw driver, L = 10 cm = 0.1 m

Now, the induce emf (E) is given as:

E = B × L × v

on substituting the values in the above equation, we get

E = 2.5 × 0.1 × 2.1

or

E = 0.525 V

Hence, the emf induced across the length of the metal shank will be 0.525 V

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A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th
alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

F=mg

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

v_x = v_0 cos \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

3 0
1 year ago
A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas
goldfiish [28.3K]
    <span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
3 0
1 year ago
What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an
Sindrei [870]

Answer:

a = 12.78 g's

Explanation:

Height reached by the object after push off is given as

H = 2.3 m

v_f^2 - v_i^2 = 2 a s

now we have

0 - v^2 = 2(-9.81)(2.3)

v = 6.72 m/s

now we know that this push last for total distance of 0.18 m

so during the push we will have

v_f^2 - v_i^2 = 2 a d

6.72^2 - 0 = 2a(0.18)

a = 125.35 m/s^2

now in terms of g = 9.81 m/s/s we have

a = \frac{125.35}{9.81} gs

a = 12.78 g's

6 0
1 year ago
Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.
marishachu [46]

Answer:

h=\frac{1}{2}\frac{v^2}{g}

Explanation:

Let's assume that an object is launched straight upward in a gravitational field. Its initial kinetic energy is given by

K=\frac{1}{2}mv^2 (1)

where m is the mass and v is the initial speed.

As the object goes higher, its kinetic energy decreases and it is converted into gravitational potential energy, since the total mechanical energy (sum of kinetic and potential energy) must remain constant:

E=K+U=const.

At the highest point of the trajectory, the speed of the object is zero (v=0), so the kinetic energy is also zero (K=0), which means that all the kinetic energy has been converted into potential energy:

U=mgh (2)

where g is the gravitational acceleration and h is the maximum height of the object.

Due to conservation of energy, we can write that (1) and (2) are equal, so:

\frac{1}{2}mv^2=mgh

from which we can derive an expression for the maximum height reached by the object

h=\frac{1}{2}\frac{v^2}{g}

5 0
1 year ago
A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
Aneli [31]
The table is almost perfect. BUT ... since she called ay negative, that means she's calling the upward direction positive-y and the downward direction negative-y. In that case, since the ball moves downward from the deck to the pavement, the change in y should be negative 3.2 m. Everything else in her table is fine. Choice-D is the good one.

Now, regarding the speed of the ball ...
How long does it take to fall 3.2 m ?
Use the formula. D = 1/2 g T^2 .

3.2 = 4.9 T^2.

T^2 = 3.2/4.9

T = √(3.2/4.9) = 0.808 second.

The ball hit the pavement 0.808 second after it rolled off the deck. So that's also the time it took to move the 0.75 m horizontally.

Speed = distance / time

Speed = 0.75 m / 0.808 second

Speed = 0.928 meter/second .
6 0
2 years ago
Read 2 more answers
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