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krok68 [10]
3 years ago
6

What happends when you slowly pull out the paper while theirs a coin on top​

Physics
2 answers:
IgorC [24]3 years ago
7 0
The correct answer is “ the coin will stay on the same place on the paper and will move along with the paper “
HACTEHA [7]3 years ago
3 0
The coin would probably slowly slide down whichever end it is closer to, weigh down the end of the paper, and fall off. It kinda depends on how you hold the paper though, because if it is directly in middle, and you hold it on both ends, it will stay on the paper.
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25

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Read 2 more answers
A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
Sophie [7]

Answer:

B = 0.37T

Explanation:

In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:

emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}       (1)

emf: induced voltage in the solenoid = 10,000V

N: turns of the solenoid = 525

ФB: magnetic flux

B: magnitude of the magnetic field = ?

A: cross-sectional area of the solenoid = π*r^2

r: radius of the cross-sectional area = 0.260m

Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s

First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90°  (quarter of a revolution)

In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):

emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}\\\\

Finally, you replace the values of the parameters to calculate B:

B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T

The strength of the magnetic field is 0.37T

7 0
3 years ago
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