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Artyom0805 [142]
3 years ago
8

A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards

the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.
Physics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

<em>b. The current in the loop always flows in a counterclockwise direction.</em>

<em></em>

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

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umka2103 [35]

Answer:

<h3>0.69</h3>

Explanation:

Using the Newtons law of motion;

\sum Fx = ma_x\\Fm - Ff = ma_x

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69

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3 years ago
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ehidna [41]

Answer:

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8 0
3 years ago
Two pulleys are fastened together to form an integral unit. At a certain instant, the indicated belt tensions act on the unit an
tresset_1 [31]

Answer:

α = 21.6 rad/s^2

Explanation:

Applying the equations of motion to determine angular acceleration of the unit,

The sum of moments about O is equal to the product of angular acceleration and moment of inertia

∑Mo = Io*α

Taking the anticlockwise direction as positive moment,

= ( -(1150) + (1400) ) * (0.5 / 2) + ( (475) - (650) ) * (0.3 / 2) - F = Io*α

= 36.5 - (2.5 N.m) =(m*ko^2)*α

NOTE: moment of inertia of the pulleys in this instance = (m*ko^2)

Hence, 33.75 = 25 * (0.25)^2 * α

Solving, α = 21.6 rad/s^2

8 0
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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
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Answer:

elastic potential

Explanation:

the rubber band has the potential to snap back when released.

7 0
3 years ago
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