Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s
Answer:
33.5 kJ
Explanation:
here there is no difference is made in the temperature. Only thing happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.
we have ice 100g =0.1 kg
Appplying Q=mL
Q= 
Q = 33 500 J
Q = 33.5 kJ