Answer:
Semiconductor. sorry i'm late
Explanation:
As we know that in order to melt the copper we need to take the temperature of copper to its melting point
So here heat required to raise the temperature of copper is given as
![Q = ms\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20ms%5CDelta%20T)
We know that
melting temperature of copper = 1085 degree C
Specific heat capacity of copper = 385 J/kg C
now we have
![Q = 3(385)(1085 - 83)](https://tex.z-dn.net/?f=Q%20%3D%203%28385%29%281085%20-%2083%29)
![Q = 1157310 J](https://tex.z-dn.net/?f=Q%20%3D%201157310%20J)
![Q = 1157.3 kJ](https://tex.z-dn.net/?f=Q%20%3D%201157.3%20kJ)
now in order to melt the copper we know the heat required is
![Q = mL](https://tex.z-dn.net/?f=Q%20%3D%20mL)
here we know that
L = 205 kJ/kg
now from above formula
![Q = 3(205) kJ](https://tex.z-dn.net/?f=Q%20%3D%203%28205%29%20kJ)
![Q = 615 kJ](https://tex.z-dn.net/?f=Q%20%3D%20615%20kJ)
now total heat required will be
![Q = 1157.3 kJ + 615 kJ](https://tex.z-dn.net/?f=Q%20%3D%201157.3%20kJ%20%2B%20615%20kJ)
![Q = 1772.3 kJ](https://tex.z-dn.net/?f=Q%20%3D%201772.3%20kJ)
As we know that
![1 Cal = 4.18 kJ](https://tex.z-dn.net/?f=1%20Cal%20%3D%204.18%20kJ)
now we have
![Q = \frac{1772.3}{4.18} = 430 KCal](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B1772.3%7D%7B4.18%7D%20%3D%20430%20KCal)
The temperature will (A) Decrease.
Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components
![x=v_0\cos20.0^\circ t+\dfrac12a_xt^2](https://tex.z-dn.net/?f=x%3Dv_0%5Ccos20.0%5E%5Ccirc%20t%2B%5Cdfrac12a_xt%5E2)
![y=v_0\sin20.0^\circ t+\dfrac12a_yt^2](https://tex.z-dn.net/?f=y%3Dv_0%5Csin20.0%5E%5Ccirc%20t%2B%5Cdfrac12a_yt%5E2)
The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components
![x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ](https://tex.z-dn.net/?f=x_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%2819%2C500%5C%2C%5Cmathrm%20m%29%5Ccos32.0%5E%5Ccirc)
![y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ](https://tex.z-dn.net/?f=y_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%2819%2C500%5C%2C%5Cmathrm%20m%29%5Csin32.0%5E%5Ccirc)
So we have enough information to solve for the components of the acceleration vector,
and
:
![x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}](https://tex.z-dn.net/?f=x_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%5Cleft%281810%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%5Ccos20.0%5E%5Ccirc%289.20%5C%2C%5Cmathrm%20s%29%2B%5Cdfrac12a_x%289.20%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%20a_x%3D21.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D)
![y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}](https://tex.z-dn.net/?f=y_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%5Cleft%281810%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%5Csin20.0%5E%5Ccirc%289.20%5C%2C%5Cmathrm%20s%29%2B%5Cdfrac12a_y%289.20%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%20a_y%3D110%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D)
The acceleration vector then has direction
where
![\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7Ba_y%7D%7Ba_x%7D%5Cimplies%5Ctheta%3D79.2%5E%5Ccirc)
I think it most likely burn since th metal tube is going to transfer so much heat