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3 years ago

5

when a body of mass 0.25kg is acted upon by a force, the velocity changes from 5m/s to 7.5m/s. Calculate the work done by the fo

rce

1 answer:

lisov135 [29]3 years ago

8 0

Answer:

Work done by the force = 1/2m×(v2 - v1)²

m = mass

v2 = final velocity

v1 = initial velocity

= 1/2(0.25)(7.5-5)²

= 1/2(0.25)(2.5)²

= 1/2(0.25)(6.25)

= 0.78J

Hope this helps.

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Given that,

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3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

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3 years ago

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Answer:

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where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

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2 years ago

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Answer:

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