A bullet of mass 0.04 kg m9ving with a speed of 90m/s enters a heavy wooden block and is stopped after a distance of 60cm. What is the average resistive force exerted bt the block on bullet.
Answers
Answered by
95
Hey
First, we have to find it's acceleration
v^2 = u^2 + 2 as
v = 0; u = 90 m/s ; s = 0.6 m (60 cm)
0 = (90)^2 + 2 × a × 0.6
0 = 8100 + 1.2 a
-8100 = 1.2 a
a = - 8100/1.2 = -6750 m/s^2
average resistive force exerted by the block on bullet =?
= mass × acceleration
= 0.04 kg × -6750 m/s^2
= 270 N (avg. only)
=======================================================
Hope this helps!
First, we have to find it's acceleration
v^2 = u^2 + 2 as
v = 0; u = 90 m/s ; s = 0.6 m (60 cm)
0 = (90)^2 + 2 × a × 0.6
0 = 8100 + 1.2 a
-8100 = 1.2 a
a = - 8100/1.2 = -6750 m/s^2
average resistive force exerted by the block on bullet =?
= mass × acceleration
= 0.04 kg × -6750 m/s^2
= 270 N (avg. only)
=======================================================
Hope this helps!
Angela1234:
crt ans dr?
Whats the meaning of "crt ans dr?"
is it correct ?
Yaa its correct
tnq ^_^
Answered by
3
Answer:270
Explanation:
Brainly.in
What is your question?
Secondary School Physics 10+5 pts
A bullet of mass 0.04 kg m9ving with a speed of 90m/s enters a heavy wooden block and is stopped after a distance of 60cm. What is the average resistive force exerted bt the block on bullet.
Report by Kranjna555oxa8gk 04.10.2017
Answers
AbhilashdhokaeHelping Hand
Know the answer? Add it here!
THE BRAINLIEST ANSWER!
Angela1234
Angela1234
Hey
First, we have to find it's acceleration
v^2 = u^2 + 2 as
v = 0; u = 90 m/s ; s = 0.6 m (60 cm)
0 = (90)^2 + 2 × a × 0.6
0 = 8100 + 1.2 a
-8100 = 1.2 a
a = - 8100/1.2 = -6750 m/s^2
average resistive force exerted by the block on bullet =?
= mass × acceleration
= 0.04 kg × -6750 m/s^2
= 270 N (avg. only)
Similar questions