Recent content by silverchain

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    Bullet stopped by sandbag: heat and temperature changes

    wow, thnx cristo, I've found the solution to question 1. Okay, i'll try to write down the formulas. question 3 i assume we'll be using the formula Pt=mc\theta so my solution, P(20s) = 1.2 x 4200 x (40-25) P(20) = 75600 P = 3780 W is this right?
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    Bullet stopped by sandbag: heat and temperature changes

    can anyone correct me if i am wrong. question 1 0.5 x 360 = 160 x 0(teta) 0(teta) = 1.125 J *why do we use 0.5 *is my answer correct? pls do tell me if I am wrong question 2 12 x 0.32 x 9.8 = 0.32 x 128 x 0(teta) 37.632 = 40.96 x teta teta = 0.91875 J * is this correct...
  3. S

    Bullet stopped by sandbag: heat and temperature changes

    ohw..thnx for the formulas...i'll try to do first. ^^
  4. S

    Bullet stopped by sandbag: heat and temperature changes

    Homework Statement 1. A bullet traveling at a speed of 360 m s-1 is stopped by a sandbag. Assuming half of the energy of the bullet becomes heat energy that is absorbed by the bullet, calculate the increase in temperature of the bullet. (Specific heat capacity of bullet = 160 J kg-1 C-1)...
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