Answer:![2.89\approx 2.9^{\circ}C/s](https://tex.z-dn.net/?f=2.89%5Capprox%202.9%5E%7B%5Ccirc%7DC%2Fs)
Explanation:
Given
![Power\left ( P\right )=150 MW](https://tex.z-dn.net/?f=Power%5Cleft%20%28%20P%5Cright%20%29%3D150%20MW)
mass of core![\left ( m\right )=1.60\times 10^5 kg](https://tex.z-dn.net/?f=%5Cleft%20%28%20m%5Cright%20%29%3D1.60%5Ctimes%2010%5E5%20kg)
Average specific heat ![\left ( C\right )=0.3349 KJ/kg^{\circ}C](https://tex.z-dn.net/?f=%5Cleft%20%28%20C%5Cright%20%29%3D0.3349%20KJ%2Fkg%5E%7B%5Ccirc%7DC)
And rate of increase of temperature =![\frac{\mathrm{d}T}{\mathrm{d} t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
Now
P=![mc\frac{\mathrm{d}T}{\mathrm{d} t}](https://tex.z-dn.net/?f=mc%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
![150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}](https://tex.z-dn.net/?f=150%5Ctimes%2010%5E6%3D1.60%5Ctimes%2010%5E5%5Ctimes%200.3349%5Ctimes%20%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
Thus ![\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Btex%5D%5Cfrac%7B1.60%5Ctimes%2010%5E5%5Ctimes%200.3349%7D%7B150%5Ctimes%2010%5E6%7D)
![\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D2.89%5Capprox%202.9%5E%7B%5Ccirc%7DC%2Fs)
Answer:
![V_1=8 V_2](https://tex.z-dn.net/?f=V_1%3D8%20V_2)
Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,
![d_2=d](https://tex.z-dn.net/?f=d_2%3Dd)
- separation distance of capacitor 1,
![d_1=2d](https://tex.z-dn.net/?f=d_1%3D2d)
- quantity of charge on capacitor 2,
![Q_2=Q](https://tex.z-dn.net/?f=Q_2%3DQ)
- quantity of charge on capacitor 1,
![Q_1=4Q](https://tex.z-dn.net/?f=Q_1%3D4Q)
We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air
![\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}](https://tex.z-dn.net/?f=%5Cepsilon_0%20%3D%208.85%5Ctimes%2010%5E%7B-12%7D%5C%2CF.m%5E%7B-1%7D)
From eq. (1)
For capacitor 2:
![C_2=\frac{k.\epsilon_0.A}{d}](https://tex.z-dn.net/?f=C_2%3D%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D)
For capacitor 1:
![C_1=\frac{k.\epsilon_0.A}{2d}](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7B2d%7D)
![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
![V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BQ%7D%7B%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%7D)
![V_2=\frac{Q.d}{k.\epsilon_0.A}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D)
& for capacitor 1:
![V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7B4Q%7D%7B%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7B2d%7D%7D)
![V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7B4Q%5Ctimes%202d%7D%7Bk.%5Cepsilon_0.A%7D)
![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)
![V_1=8 V_2](https://tex.z-dn.net/?f=V_1%3D8%20V_2)
Answer:
The voltage will be 0.0125V
Explanation:
See the picture attached
by an echo meter
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#Genius kudi
Conductivity is the property of matter in which a substance can transfer heat or electricity