a bomb is kept stationary at a point it suddenly explodes into two fragments of mass 1 gram and 3 grand. total kinetic energy of the fragment is 6.4 * 10^4j. what is the kinetic energy of the smaller fragment.
Solve properly.
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✌♥✌hey mate here is your answer ✌♥✌
⛄total kinetic energy=6.4×10⁴j♥
⛄kinetic energy of smaller fragment=♥
⛄ 1/2 ×1×10^-3×v1²♥
⛄and that of bigger fragment=♥
⛄ 1/2 ×3×10^-3×v2²♥
⛄the velocity and momentum will be opposite to each other♥
⛄so 1×10^-3×v1=3×10^-3×v2♥
⛄ v1=3v2♥
⛄so♥
⛄ 1/2 ×1×10^-3×v1²+1/2 ×3×10^-3×v2²=6.4×10⁴
⛄1/2 ×1×10^-3×9v2²+1/2 ×3×10^-3×v2²=6.4×10⁴
⛄1/2×10^-3(12v2²)=6.4×10⁴♥
⛄v2²=64/6 ×10^6♥
⛄so v1²=9×64/6 ×10^6♥
⛄ =96×10^6♥
⛄ v1=4√6×10³m/s♥
⛄so♥
⛄kinetic energy of smaller fragment=♥
⛄ 1/2 ×1×10^-3×v1²♥
⛄=1/2 ×10^-3×96×106♥
⛄=4.8×10⁴joules♥
⛄hope it will help you ♥
✌♥✌mark me brainliest ✌ ♥✌
⛄total kinetic energy=6.4×10⁴j♥
⛄kinetic energy of smaller fragment=♥
⛄ 1/2 ×1×10^-3×v1²♥
⛄and that of bigger fragment=♥
⛄ 1/2 ×3×10^-3×v2²♥
⛄the velocity and momentum will be opposite to each other♥
⛄so 1×10^-3×v1=3×10^-3×v2♥
⛄ v1=3v2♥
⛄so♥
⛄ 1/2 ×1×10^-3×v1²+1/2 ×3×10^-3×v2²=6.4×10⁴
⛄1/2 ×1×10^-3×9v2²+1/2 ×3×10^-3×v2²=6.4×10⁴
⛄1/2×10^-3(12v2²)=6.4×10⁴♥
⛄v2²=64/6 ×10^6♥
⛄so v1²=9×64/6 ×10^6♥
⛄ =96×10^6♥
⛄ v1=4√6×10³m/s♥
⛄so♥
⛄kinetic energy of smaller fragment=♥
⛄ 1/2 ×1×10^-3×v1²♥
⛄=1/2 ×10^-3×96×106♥
⛄=4.8×10⁴joules♥
⛄hope it will help you ♥
✌♥✌mark me brainliest ✌ ♥✌
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