Answer:
mercury and alcohol
ii) used to test temperatures
Answer:
Weight
Explanation:
Weight is the downward pull on an object due to gravity.
For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.
It is 10.20 m from the ground.
<u>Explanation:</u>
<u>Given:</u>
m = 0.5 kg
PE = 50 J
We know that the Potential energy is calculated by the formula:

where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.
PE is the Potential Energy.
Potential Energy is the amount of energy stored when an object is stationary.
Here, if we substitute the values in the formula, we get

50 = 0.5 × 9.8 × h
50 = 4.9 × h

h = 10.20 m
3200÷0.22= 145.4545...N
(it is an infinite decimal)
Answer:

Explanation:
Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

now we can say that


now electric potential is given as


now work done to bring a small charge from infinite to the surface of this sphere is given as


here we know that

now the total energy of the sphere is given as



