A circular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing downward, and is released from rest. What is the direction of the induced current in the coil, as viewed from above, as the magnet approaches the coil in free fall?a. clockwise
b. counterclockwise
c. There is no induced current in the coil.

Answers

Answer 1

Answer:

Answer:

Option B

Explanation:

As per the Lenz’s law of electromagnetism the current induced in a conductor due to any change has a tendency to oppose the change which is causing this induces current.

Thus, when a constant magnetic field with an electric circuit is varied, it produces and induced current which flow in a direction such that its sets a magnetic field that tries to restore the flux

Hence, option B is correct

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A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels640 m in 40 s. For the 40 s interval, find (a) the average velocity, (b) the final velocity, and (c) theacceleration.
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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 J
B) 5 J
C) 50 J
D) 1 J
E) 10 J

Answers

Answer:

option C

Explanation:

given,

Force on the object = 10 N

distance of push = 5 m

Work done = ?

we know,

work done is equal to Force into displacement.

W = F . s

W = 10 x 5

W = 50 J

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

Final answer:

The work done on an object when a force of 10 N pushes it 5 m is 50 Joules, calculated by multiplying the force and the displacement. So, the correct option is C.

Explanation:

The question is asking about work, which in physics is the result of a force causing a displacement. The formula for work is defined as the product of the force (in Newtons) and the displacement (in meters) the force causes. If a force of 10 N pushes an object a distance of 5 m, the work done is calculated by multiplying the force and the displacement (10 N * 5 m), yielding 50 Joules of work.

Therefore, the correct answer is 50 J (C).

Learn more about Work here:

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If you know the distance of an earthquake epicenter from three seismic stations, how can you find the exact location of the epicenter of the earthquake.

Answers

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

Answers

Answer

given,

Tension of string is F

velocity is increased and the radius is not changed.

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone

= $(mv^2)/(r)$

now centrifugal force is balanced by tension

T = $(mv^2)/(r)$

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?

Answers

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations (1 minute = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

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