R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%
Answer:
The change of the volume of the device during this cooling is ![14.3\times10^{-3}\ m^3](https://tex.z-dn.net/?f=14.3%5Ctimes10%5E%7B-3%7D%5C%20m%5E3)
Explanation:
Given that,
Mass of oxygen = 10 g
Pressure = 20 kPa
Initial temperature = 110°C
Final temperature = 0°C
We need to calculate the change of the volume of the device during this cooling
Using formula of change volume
![\Delta V=V_{2}-V_{1}](https://tex.z-dn.net/?f=%5CDelta%20V%3DV_%7B2%7D-V_%7B1%7D)
![\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cdfrac%7BmR%7D%7BP%7D%28T_%7B2%7D-T_%7B1%7D%29)
Put the value into the formula
![\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cdfrac%7B0.3125%5Ctimes0.0821%7D%7B2.0265%5Ctimes10%5E%7B9%7D%7D%28383-273%29)
![\Delta V=14.297\ L](https://tex.z-dn.net/?f=%5CDelta%20V%3D14.297%5C%20L)
![\Delta V=14.3\times10^{-3}\ m^3](https://tex.z-dn.net/?f=%5CDelta%20V%3D14.3%5Ctimes10%5E%7B-3%7D%5C%20m%5E3)
Hence, The change of the volume of the device during this cooling is ![14.3\times10^{-3}\ m^3](https://tex.z-dn.net/?f=14.3%5Ctimes10%5E%7B-3%7D%5C%20m%5E3)
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)
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