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A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mass melts be m,specific heat S , initial temperature 25 degree Celsius, melting point 475 degree Celsius and latent heat L.then v is given by?
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A bullet moving with a uniform velocity v, stops suddenly after hittin...
1/2mv^2=ms450+ml
Solve for v u get v=(2l+900s)ans.
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A bullet moving with a uniform velocity v, stops suddenly after hittin...
Given data:
- Initial velocity of the bullet, $v$
- Mass of the target, $m$
- Specific heat of the material, $S$
- Initial temperature, $T_i = 25^\circ C$
- Melting point, $T_m = 475^\circ C$
- Latent heat of fusion, $L$

Finding the final temperature:
- The bullet stops suddenly after hitting the target, transferring all its kinetic energy into heat energy.
- The heat energy required to raise the temperature of the mass from $T_i$ to $T_m$ is given by $Q = m \cdot S \cdot (T_m - T_i)$.
- The heat energy required to melt the mass at $T_m$ is given by $Q = m \cdot L$.
- Equating the two heat energies, we get $m \cdot S \cdot (T_m - T_i) = m \cdot L$.
- Solving for $T_m$, we get $T_m = \frac{L}{S} + T_i$.

Calculating the velocity:
- The final kinetic energy of the bullet is zero.
- The initial kinetic energy of the bullet is $\frac{1}{2} m v^2$.
- The work done in stopping the bullet is equal to the initial kinetic energy.
- The work done is given by $Q = \frac{1}{2} m v^2$.
- Substituting the value of $Q$ from above, we get $\frac{1}{2} m v^2 = m \cdot S \cdot (T_m - T_i)$.
- Substituting the value of $T_m$, we get $\frac{1}{2} v^2 = \frac{L}{2S}$.
- Solving for $v$, we get $v = \sqrt{\frac{L}{S}}$.
Therefore, the velocity of the bullet, $v$, is given by $\sqrt{\frac{L}{S}}$.
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A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mass melts be m,specific heat S , initial temperature 25 degree Celsius, melting point 475 degree Celsius and latent heat L.then v is given by?
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