The momentum of the ball when it hits the ground is 4.89 kg.m/s.
The given parameters;
- <em>mass of the baseball, m = 0.145 kg</em>
- <em>height of fall of the ball, h = 58 m</em>
The final velocity of the ball when it hits the ground is calculated as follows;
The momentum of the ball when it hits the ground is calculated as follows;
P = mv
P = 0.145 x 33.72
P = 4.89 kg.m/s
Thus, the momentum of the ball when it hits the ground is 4.89 kg.m/s.
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Answer:
(a) 693.12 torr
(b) 68.5 kilopascals
(c) 0.862 atmosphere
(d) 1.306 atmospheres
(e) 36.74 psi
Explanation:
(a) 0.912 atm = 0.912 atm × 760 torr/1 atm = 693.12 torr
(b) 0.685 bar = 0.685 bar × 100 kPa/1 bar = 68.5 kPa
(c) 655 mmHg = 655 mmHg × 1 atm/760 mmHg = 0.862 atm
(d) 1.323×10^5 Pa = 1.323×10^5 Pa × 1 atm/1.01325×10^5 Pa = 1.306 atm
(e) 2.50 atm = 2.50 atm × 14.696 psi/1 atm = 36.74 psi
Answer:
22m/s
Explanation:
Mass, m=60 kg
Force constant, k=1300N/m
Restoring force, Fx=6500 N
Average friction force, f=50 N
Length of barrel, l=5m
y=2.5 m
Initial velocity, u=0
Substitute the values
m
Work done due to friction force
We have 
Substitute the values
Initial kinetic energy, Ki=0
Initial gravitational energy, 
\
Initial elastic potential energy
Final elastic energy,
Final kinetic energy, 
Final gravitational energy, 
Final gravitational energy, 
Using work-energy theorem
Substitute the values
Answer:
75 grams
real is 85
Explanation:
you say you where 10 grams off.
