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A 20g bullet moving at 200 meter per second hits a bag of sand and comes to rest in 0.011secs.What is the momentum of the bulle...Asked by Anonymous
A 20g bullet moving at 200ms hit a bag of sand and comes to rest in 0.011 s what is the momentum of the bullet just before hitting the bag? find the average force that stopped the bullet
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Answered by
Hi yah
momentum = mass * velocity ... kg•m/s
acceleration = v / t = 200 m/s / 11 ms ... m/s^2
force = mass * acceleration = .02 kg * (200 m/s / 11 ms) ... N
acceleration = v / t = 200 m/s / 11 ms ... m/s^2
force = mass * acceleration = .02 kg * (200 m/s / 11 ms) ... N
Answered by
Muhamm@d
Answer the question.a 20g bullet moving at 200ms1 hits a bag of sand and come to rest in 0.011s . what is the momentum of the bullet just before hitting the bag?find the average force that stopped the bullet.
Answered by
Brian meme
Have not seen my answer
Answered by
bolu
this is rubbish
Answered by
Dora
I want my answer
Answered by
Hmmmm
Where is the answer
Waste of time though🙄
Waste of time though🙄
Answered by
Emmanuel
Momentum=mass x velocity
0.02 x 200 =4
0.02 x 200 =4
Answered by
Anonymous
Olodo
Answered by
Similoluwa
Since p=MV,then 20*200=4000
Answered by
Esther
Pls i need an answer for the question
Answered by
Damsel
Like s
Answered by
OMALE ESTHER
PLS IS THIS RIGHT
Answered by
Frank
He is right by his steps
Answered by
Marcus
Momentum=p(mv)
So p = 0.02×200
P = 4kgms
Average force
F= change in p
F= 4/0.011
F= 363.6N
So p = 0.02×200
P = 4kgms
Average force
F= change in p
F= 4/0.011
F= 363.6N
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