A 5.0 g silver-bullet is fired into a fence post. The initial speed of the bulle

Question

A 5.0 g silver-bullet is fired into a fence post. The initial speed of the bullet is 290 m/s, and when it comes to rest, half its kinetic energy goes into heating the bullet. How much does the bullet's temperature increase?

A 5.0 g silver-bullet is fired into a fence post. The initial speed of the bullet is 290 m/s, and when it comes to rest, half its kinetic energy goes into heating the bullet. How much does the bullet's temperature increase? 164.2578 Your response differs from the correct answer by more than 10%. Double check your calculations. K

Explanation / Answer

  mass = 5g = 0.005kg

Initial K.E = (1/2)m v^2

=(1/2) x 0.005 x 290^2

= 210.5 J

When it comes to rest, half of it's K.E is is transformed into heat energy.

---> Heat energy = (1/2) x K.E

= (1/2) x210.5

= 105.125 J

Heat energy = mc[delta(t)] ------------> delta(t) = temperature increase/change

0.005 x 230 x delta(t) = 105.125

delta(t) = 91.42K or 91.42 degree Celsius.

1 pound = 0.454kg, so 13lb = 5.902 kg
1 foot = 0.3048m, 1.3feet = 0.397m
Assuming you mean '140Cal' (with a capital 'C' not a small 'c')
1kcal (=1Cal) = 4184J,so 110Cal = 585760 J
Take g=9.81m/s^2 (9.81N/kg)

Gravitational potential energy needed for 1 rep (raise mass) = mgh=5.902x9.81x0.397= 22.99 J

No. of rep = 585760/22.99 = 25478.91 approx

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