Answer:
Explanation:
Calories to be burnt = 3500 - 2500 = 1000 Cals .
Efficiency of conversion to mechanical work is 25 % .
Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.
4000 Cals = 4.2 x 4000 = 16800 J .
Work done in one jump = kinetic energy while jumping
= 1/2 m v²
= .5 x 70 x 3.3²
= 381.15 J .
Number of jumps required = 16800 / 381.15
= 44 .
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Answer:
KE = 0.162 KJ
Explanation:
given,
mass of bullet (m)= 20 g = 0.02 Kg
speed of the bullet (u)= 1000 m/s
mass of block(M) = 1 Kg
velocity of bullet after collision (v)= 100 m/s
kinetic energy = ?
using conservation of momentum
m u = m v + M V
0.02 x 1000 = 0.02 x 100 + 1 x V
20 = 2 + V
V = 18 m/s
now,
Kinetic energy of the block
KE = 162 J
KE = 0.162 KJ
Answer:
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Explanation:
so confusing
