A bullet loses 25% of its initial kinectic energy after penetrating through a distance of 2 cm in a target .the further distance it travels before coming to rest is
Answers
Answered by
15
Acceleration = u² / (2S)
= u² / (2 × 0.02)
= u² / 0.04
Loss in kinetic energy = 0.5m(u² - v²)
(25/100) × 0.5 mu² = 0.5m(u² - v²)
u²/4 = (u² - v²)
v = (u√3)/2
Further distance = v² / (2a)
= (3u²/4) / [2 × (u² / 0.04)]
= (3 × 0.04) / (4 × 2)
= 0.015 m
= 1.5 cm
Further distance bullet travels is 1.5 cm
= u² / (2 × 0.02)
= u² / 0.04
Loss in kinetic energy = 0.5m(u² - v²)
(25/100) × 0.5 mu² = 0.5m(u² - v²)
u²/4 = (u² - v²)
v = (u√3)/2
Further distance = v² / (2a)
= (3u²/4) / [2 × (u² / 0.04)]
= (3 × 0.04) / (4 × 2)
= 0.015 m
= 1.5 cm
Further distance bullet travels is 1.5 cm
Answered by
4
Answer: the answer is 6cm
Explanation:
The answer is given in the attachment
Please mark me as brainliest
Attachments:
Similar questions