A lead bullet strikes a target with a velocity of `500 ms^(-1)` and the bullet falls dead. Calculate the rise in temperature of the bullet assuming that 40% of the heat produced is used in heating the bullet. Given specific heat of lead `= 0.03 cal g^(-1).^(@)C^(-1), J = 4.2 J cal^(-1)`.

A lead bullet strikes a target with velocity of 480 m/s. if the bullet falls dead, then the rise in temperature of bullet is, (Assuming that heat produced is equally shared between the bullet and target). (J=4.2xx10^(3)J//kcal,C=0.03kcal//kgK )

A lead bullet weighing 20 g and moving with a velocity of 100 ms^(-1) comes to rest in a fixed block of wood. Calculate the heat developed and the rise in temperature of the bullet assuming that the half of the heat is absorbed by the bulley. The specific heat of lead is 0.03 cal g^(-1).^(@)C^(-1) .

A bullet of mass 20xx10^(-3)kg enters into a fixed block of wood with a velocity of 100ms^(-1) and is brought to rest in the wood. Calclate the rise in temperature of te bullet if two -third of the heat produced is absorbed by the bullet. (Specific heat capacity of lead =32calkg^(-1)K^-1) and J=4.2J//"callorie" )

A 30 g bullet leaves a rifle with a velocity of 300 ms^(-1) and the rifle recoils with a velocity of 0.60ms^(-1) . Find the mass of the rifle.

A lead bullet of mass 50g moving with a velocity of 100ms^(-1) strikes a target. If 40% of the heat produced is retained by the bullet, calculate the rise in its temperature. Specific heat of lead = 0.03 cal g^(-1)K^(-1) and J = 4.2 J cal^(-1)

A lead bullet strikes against a steel plate with a velocity 200ms^-1 . If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, thent he rise in temperature of the bullet is (specific heat capacity of lead =125Jkg^-1K^-1 )

With what velocity must a lead bullet at 50^(@)C strike against an obstacle in order that the heat produced by the arrest of the motion, is sufficient of melt it, assuming that all the heat produced remains within the bullet? (Specific heat capacity of lead =31cal kg^(-1)K^(-1) , melting point of lead =335^(@)C , specific latent heat capacity of lead =5370 cal kg^(-1)K^(-1) and J=4.2 joules cal^(-1) )

The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall ? [Given 1 cal=4.2 J and specific heat of water =1 cal g ^(-1) ""^(@)C ^(-1) ]