Question

The belt of an electrostatic generator is 50 cm wide and travels at 30 cm/s . The belt carries charge into the sphere at a rate which corresponds to $10^{-4}$ ampere. What is surface density of charge on the belt?

Answer 1

Answer is in the attachment :(

Answer 2

ANSWER:

The upward acceleration of the lift = 3g m/s².

GIVEN:

A simple pendulum is suspended from the ceiling of a lift.

When the lift is at rest its time period is T.

New time period of the pendulum = T/2.

TO FIND:

The upward acceleration of the lift.

EXPLANATION:

$\pink\bigstar\green{ \boxed{\large{\bold{\orange{T = 2\pi\sqrt{\dfrac{l}{g}}}}}}} \\ \\ \\ \sf T' = 2 \pi \sqrt{\dfrac{l}{a} } \\ \\ \\ \sf W.K.T. \ T' = \dfrac{T}{2} \\ \\ \\ \sf \dfrac{T}{2} = 2 \pi \sqrt{\dfrac{l}{a} } \\ \\ \\ \sf \dfrac{ 2 \pi}{2} \sqrt{\dfrac{l}{g} } = 2 \pi \sqrt{\dfrac{l}{a} } \\ \\ \\ \sf \dfrac{ 1}{2} \sqrt{\dfrac{1}{g} } = \sqrt{\dfrac{1}{a} } \\ \\ \\ \sf \dfrac{ 1}{4g} = \dfrac{1}{a} \\ \\ \\ \sf a_{net} = 4g \\ \\ \\ \sf a_{net} = g + a_{upwards} \\ \\ \\ \sf 4g = g + a_{upwards} \\ \\ \\ \sf a_{upwards} = 3g$

•°• The upward acceleration of the lift = 3g m/s².