A 12 gram bullet is accelerated from rest to a speed of 700 m/s as it travels 20 cm in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force?
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Answers
Answered by
31
v² - u² = 2 a s
a = (700² - 0² )/ 2 * 0.20 = 12.25 * 10^5 m/sec
force = ma = 14700 N as m = 0.012 kg
a = (700² - 0² )/ 2 * 0.20 = 12.25 * 10^5 m/sec
force = ma = 14700 N as m = 0.012 kg
shanmu54321:
the answer isn't correct. I already tried your method
check now. a calculation mistake only it was. do you understand the formula.
awesome man!!!!!! TNX
Answered by
50
m = 12 g = 12/1000 kg = 0.012 kg,
u = 0 m/s
v = 700 m/s
S (distance) = 20 cm = 20/100 m = 0.2 m
Formula used:
v² = u² + 2aS
(700)² = (0)² + 2. a. (0.2)
490000 = 0.4 a
a = 490000/0.4
a = 1225000 m/s²
Force= 1225000 * 0.012
= 147000 newton
u = 0 m/s
v = 700 m/s
S (distance) = 20 cm = 20/100 m = 0.2 m
Formula used:
v² = u² + 2aS
(700)² = (0)² + 2. a. (0.2)
490000 = 0.4 a
a = 490000/0.4
a = 1225000 m/s²
Force= 1225000 * 0.012
= 147000 newton
the answer isn't correct. I already tried your method
Thanks for the effort
Thanks to the previous user.. I completely ignored the force part
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