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 of mass 10 g travelling at 300 m//s strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead = 150 J//kg^(@)C )

A piece of lead is dropped to the ground from a height of 200 m. What will be the increase in its temperature? Specific heat of lead = 0.03 cal cdot g^(-1) cdot^(@)C^(-1), J = 4.2 xx 10^(7) erg cdot cal^(-1)

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 an iron wall with a velocity of 400 ms^( -1). If the bullet falls dead and the heat produced is equally shared between bullet and the wall find the rise in temperature of the bullet. Given, specific heat of bullet = 125.6 J kg ^(-1) K ^(-1).

A lead ball moving with a velocity v strikes a wall and stops. If 50% of its energy is onverted into heat. The increase in temperature is (Specific heat of lead is S)

A lead bullet of 10g travelling at 300m//s strikes against a block of wood and comes to rest. Assuming 50% heat is absorbed by the bullet, the increase in its temperature is (sp-heat of lead is 150J//Kg-K )

A lead bullet of 10g travelling at 300m//s strikes against a block of wood and comes to rest. Assuming 50% heat is absorbed by the bullet, the increase in its temperature is (sp-heat of lead is 150J//Kg-K )