Answer:
0.1667 m/s
Explanation:
m1V1 + m2V2 = m1V3 + m2V4
0.01 = ( 0.0075) + (0.015 * V4)
V4 = (0.01 - 0.0075) / (0.015)
V4= 0.1667
Answer:
The average acceleration of the bearings is 
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval 
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration

Put the value into the formula



Hence,The average acceleration of the bearings is 
Acceleration due to gravity.
Answer:
Specific heat at constant pressure is = 1.005 kJ/kg.K
Specific heat at constant volume is = 0.718 kJ/kg.K
Explanation:
given data
temperature T1 = 50°C
temperature T2 = 80°C
solution
we know energy require to heat the air is express as
for constant pressure and volume
Q = m × c × ΔT ........................1
here m is mass of the gas and c is specific heat of the gas and Δ
T is change in temperature of the gas
here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.
and here at constant pressure Specific heat is greater than the specific heat at constant volume,
so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is
Specific heat at constant pressure is = 1.005 kJ/kg.K
and
Specific heat at constant volume is = 0.718 kJ/kg.K