If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So

Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar
Answer:
The time to boil the water is 877 s
Explanation:
The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added
V = i (r + R)
We replace we calculate
r + R = V / i
R = v / i - r
R = 10/12 -0.04
R = 0.793 Ω
We calculate the power supplied
P = V i = I² R
P = 12² 0.793
P = 114 W
This is the power dissipated in the external resistance
We use the relationship, that power is work per unit of time and that work is the variation of energy
P = E / t
t = E / P
t = 100 10³/114
t = 877 s
The time to boil the water is 877 s