Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
K.E.= 1/2 x MV^2 = 1/2 x 40(kg) x (25x25) =12500J
Answer:every force has an equal and oppisite reaction. meeaning if you punch a wall wwith 50lbs of force the wall pushes back on your hand with 50 lbs of force
Explanation:
Answer:
A. Final pressure P2
P2/P1 = (T2/T1)^n/n-1
P1 = 4bar
T1 = 438K
T2 = 300K
Polytropic index, n, = 1.3
P2 = 4 (300/438)^1.3/1.3-1
P2 = 4 (300/438)^4.333
P2 = 4 * 0.19400
P2 = 0.776bar.
B. The work done is;
W = mR/ n-1 (T1 -T2)
Where, R = 0.1889kJ/kg.K, m = 1
W = 1 * 0.1889/ 1.3-1 * (438-300)
W = 86.89kJ/kg.
C. The heat transfer, Q
Q = W + ΔU
Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk
Q = 86.89 + 1 * 0.743 (300-438)
Q = 86.89 + (-102.534)
Q = -15.644kJ/K
Q = 15.64kJ/K
