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NemiM [27]
2 years ago
13

Can someone solve it?

Physics
1 answer:
True [87]2 years ago
3 0

Answer:A

Explanation:

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A 0.43 m long and 0.43 m wide loop is moved at a constant velocity of 0.15 m/s into a perpendicular constant magnetic field of 0
olya-2409 [2.1K]

Answer:

The magnitude of the induced voltage in the loop is 20 mV.

Explanation:

given;

length of loop, L = 0.43 m

width of loop,w = 0.43 m

velocity of moved loop, v = 0.15m/s

magnetic field strength,B = 0.31 T

To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;

magnitude induced E.M.F = BLv

magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV

Therefore, the magnitude of the induced voltage in the loop is 20 mV.

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3 years ago
What is the frequency of a wave that has a period of vibration of 2 seconds?
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Answer:

The answer is 0.5 Hz

Explanation:

Its pretty easy to get the answer. One hertz (Hz) is equal to one cycle or period per second. So, just divide the period by the number of seconds.

1 period/2 secs = 1/2 Hz or 0.5 Hz

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Seasonal changes bring about scenes like this one. Plant cells respond to changes in _________________ and as a result, photosyn
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A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

Because tension is equal to centripetal force

Substitute the values

T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

W=\frac{1}{2}m(v^2-v^2_0)

W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

W=0.31 J

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3 years ago
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