Asked by Anonymous
If the rifle is stopped by the hunter’s shoulder in a distance of 3.16 cm, what is the magnitude of the average force exerted on the shoulder by the rifle?
Answer in units of N.
mass of bullet= 0.0137 kg
velocity of bullet= 546 m/s to the right
mass of rifle= 3.82 kg
recoil speed of the rifle as the bullet leaves the rifle= 1.958167539 m/s
Answer in units of N.
mass of bullet= 0.0137 kg
velocity of bullet= 546 m/s to the right
mass of rifle= 3.82 kg
recoil speed of the rifle as the bullet leaves the rifle= 1.958167539 m/s
Answers
Answered by
Bob
1. Change the distance from cm to meters by dividing by 100.
3.16cm -> 0.0316m
2. Use this formula to find time.
t= 2(x-displacement-)/V(speed)
2 • 0.0316m/1.958167539 = 0.032275s
3. Use this formula for Force F=mv/t
3.82kg • 1.958167539/ 0.032275 = 231.765N
The answer is 231.765N
3.16cm -> 0.0316m
2. Use this formula to find time.
t= 2(x-displacement-)/V(speed)
2 • 0.0316m/1.958167539 = 0.032275s
3. Use this formula for Force F=mv/t
3.82kg • 1.958167539/ 0.032275 = 231.765N
The answer is 231.765N
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