A bullet of mass 25 g moving with a velocity of 200 m/s is stopped within 5 cm of the target. the average resistance offered by the target is
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mark branliest
mark branliest
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Hey !
Given:
Mass of bullet m = 25 g = 0.025 kg
Solution:
Initial velocity of bullet,
u = 200 m/s
Final velocity, v = 0 and
distance, s = 5 cm = 0.05 m
Relation for the acceleration
v² = u² - 2as
(or) 0 = (200)² - 2a × 0.05
(or) a = (200)²/2 × 0.05
(or) = 400000 m/sec²
∴ Average resistance offered by the target,
F = m.a
= 0.025 × 400000
= 10000 N = 10 kN
GOOD LUCK !
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