A bullet travles horizontaly at 865m/s and enters a tank containing 13.5kg of wa

Question

A bullet travles horizontaly at 865m/s and enters a tank containing 13.5kg of waternand exits the tank at 534 m/s. ¿Whats the max. Temperature the water could reach as a result of this event?
Need whole procedure Will rate thumbs up correct awnser :) A bullet travles horizontaly at 865m/s and enters a tank containing 13.5kg of waternand exits the tank at 534 m/s. ¿Whats the max. Temperature the water could reach as a result of this event? A bullet travles horizontaly at 865m/s and enters a tank containing 13.5kg of waternand exits the tank at 534 m/s. ¿Whats the max. Temperature the water could reach as a result of this event?
Need whole procedure Will rate thumbs up correct awnser :)

Explanation / Answer

KE is lost during this process, this difference in KE will be absorbed by water that will raise its temp.

KE_lost = m (vi^2 - vf^2) /2

= m (865^2 - 534^2) /2 = 231534.5 m

and Water, Q = m C deltaT  

231534.5m = (13.5)(4186)(deltaT)

delta(T) = 4.097 m

value of m (mass of bullet) is not given. Put m in kg.

suppose mass of bullet is 15 grams.

then m = 15 x 10^-3 kg

delta(T) = (15 x 10^-3)(4.097) = 0.061 deg C

  

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