A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The in
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Answer:
(a). The required is 1871.2 J.
(b). The speed the lead bullet is 395 m/s.
(c). The 20% of energy must have gone into collision between steel plate and bullet.
Explanation:
Given that,
Mass of bullet = 0.030 kg
Temperature = 20°C
(a). We need to heat required to increase the temperature of the lead bullet and melt it
Using formula of heat
Where, m = mass of lead bullet
S = specific heat
L = latent heat
T = Temperature
Put the value into the formula
The required is 1871.2 J.
(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it
We need to calculate the energy
We need to calculate the speed the lead bullet
Using formula of speed
The speed the lead bullet is 395 m/s.
(c). The 20% of energy must have gone into collision between steel plate and bullet.
Hence, This is the required solution.