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igor_vitrenko [27]
3 years ago
8

Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

val (on Earth) the Moon clock:
A) (9.8/1.6)h
B) 1 h
C) the square root of 9.8/1.6 h
D) (1.6/9.8)h
E) the square root of 1.6/9.8 h
Physics
1 answer:
satela [25.4K]3 years ago
7 0

The time-period of a simple pendulum is

<em>Time =  2 π √(length/grav-accel)</em>

After unraveling the question, then completing it, and working out what I <em>believe</em> it's trying to ask, the choice that correctly answers the question that I have invented is <em>choice-E</em> .

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What is the net force acting on the object above?
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As neutron is neutral,there will be no change in the electrostatic forces as there is no increment of charge.but when the neutron number is increased at that time the binding energy per nucleon is decreased which in turn affects the nuclear force which is a  short range force.the nucleus becomes heavy and unstable which undergoes decay.

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Answer:

<h3>(A) The width x = 0.24 \times 10^{-3} m</h3><h3>(B) The new width is 1.32 \times 10^{-3} m</h3>

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We know that \theta is proportional to the x and inversely proportional to the D

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