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3 years ago

When a conducting plate swings through a magnetic field, which is true?

Physics

2 answers:

Nookie1986 [14]3 years ago

7 0

Answer:

Energy is transferred from mechanical energy to thermal energy. (A)

Explanation:

As the conducting plate is swinging, we know it posseses some form of mechanical energy (either K or U). Because the plate is conducting, charge is transfered, yielding energy transfer to thermal.

VLD [36.1K]3 years ago

4 0

Answer:

Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy ( D )

Explanation:

when conducting plate swings through a magnetic field, energy is transferred from mechanical energy to thermal energy and then back to mechanical energy and this is due to the induction of eddy currents during the process of that the conducting plate swings through the magnetic field.

An eddy current is induced in conductor when the magnetic field in which the conductor is located varies.

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RoseWind [281]

Answer:

4. It is the force of the road on the tires (an external force) that stops the car.

Explanation:

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You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.

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3 years ago

The value of g is greater at the poles than at the equator why

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Answer:

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3 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

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Initial momentum

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

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$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

Average Force

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

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3 years ago

Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro

aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

$S = \frac{Q}{T}$

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

$\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}$

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy $\Delta S_{hot}$ is smaller than $\Delta S_{cold}$