Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.
To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
The extension of the spring due to the weight of the object on Moon is a value of 
, then
Recall that gravity on the moon is a sixth of Earth's gravity.
We have that the displacement at the earth was 
, then
Therefore the displacement of the mass on the spring on Moon is 0.05m
A funnel cloud is a funnel-shaped cloud of condensed water droplets. They usually appear with a rotating column of air. These extend from the bottom of a cloud that does not touch the ground or a water surface.
