Answer:
The temperature of cold reservoir should be 246.818 K for efficiency of 35%
Explanation:
In first case we have given efficiency of Carnot engine = 26 % = 0.26
Temperature of cold reservoir ![T_L=281K](https://tex.z-dn.net/?f=T_L%3D281K)
We know that efficiency of Carnot engine is given by
![\eta =1-\frac{T_L}{T_H}](https://tex.z-dn.net/?f=%5Ceta%20%3D1-%5Cfrac%7BT_L%7D%7BT_H%7D)
![0.26 =1-\frac{281}{T_H}](https://tex.z-dn.net/?f=0.26%20%3D1-%5Cfrac%7B281%7D%7BT_H%7D)
![T_H=379.72K](https://tex.z-dn.net/?f=T_H%3D379.72K)
For second Carnot engine efficiency is given as 35% = 0.35
And temperature of hot reservoir is same so ![T_H=379.72K](https://tex.z-dn.net/?f=T_H%3D379.72K)
So ![0.35=1-\frac{T_L}{379.72}](https://tex.z-dn.net/?f=0.35%3D1-%5Cfrac%7BT_L%7D%7B379.72%7D)
![T_L=246.818K](https://tex.z-dn.net/?f=T_L%3D246.818K)
So the temperature of cold reservoir should be 246.818 K for efficiency of 35%
Answer:
Part a)
![\lambda = 300 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20300%20m)
Part b)
![E = 2.7 N/C](https://tex.z-dn.net/?f=E%20%3D%202.7%20N%2FC)
Part c)
![I = 9.68 \times 10^{-3} W/m^2](https://tex.z-dn.net/?f=I%20%3D%209.68%20%5Ctimes%2010%5E%7B-3%7D%20W%2Fm%5E2)
![P = 3.22 \times 10^{-11} N/m^2](https://tex.z-dn.net/?f=P%20%3D%203.22%20%5Ctimes%2010%5E%7B-11%7D%20N%2Fm%5E2)
Explanation:
Part a)
As we know that frequency = 1 MHz
speed of electromagnetic wave is same as speed of light
So the wavelength is given as
![\lambda = \frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bc%7D%7Bf%7D)
![\lambda = \frac{3\times 10^8}{1\times 10^6}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B3%5Ctimes%2010%5E8%7D%7B1%5Ctimes%2010%5E6%7D)
![\lambda = 300 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20300%20m)
Part b)
As we know the relation between electric field and magnetic field
![E = Bc](https://tex.z-dn.net/?f=E%20%3D%20Bc)
![E = (9 \times 10^{-9})(3\times 10^8)](https://tex.z-dn.net/?f=E%20%3D%20%289%20%5Ctimes%2010%5E%7B-9%7D%29%283%5Ctimes%2010%5E8%29)
![E = 2.7 N/C](https://tex.z-dn.net/?f=E%20%3D%202.7%20N%2FC)
Part c)
Intensity of wave is given as
![I = \frac{1}{2}\epsilon_0E^2c](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cepsilon_0E%5E2c)
![I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%288.85%20%5Ctimes%2010%5E%7B-12%7D%29%282.7%29%5E2%283%5Ctimes%2010%5E8%29)
![I = 9.68 \times 10^{-3} W/m^2](https://tex.z-dn.net/?f=I%20%3D%209.68%20%5Ctimes%2010%5E%7B-3%7D%20W%2Fm%5E2)
Pressure is defined as ratio of intensity and speed
![P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BI%7D%7Bc%7D%20%3D%20%5Cfrac%7B9.68%5Ctimes%2010%5E%7B-3%7D%7D%7B3%5Ctimes%2010%5E8%7D)
![P = 3.22 \times 10^{-11} N/m^2](https://tex.z-dn.net/?f=P%20%3D%203.22%20%5Ctimes%2010%5E%7B-11%7D%20N%2Fm%5E2)
Answer:
29223.6J
Explanation:
Given parameters:
Mass of Piano = 852kg
Height of lifting = 3.5m
Unknown:
Gravitational potential energy = ?
Solution:
The gravitational potential energy of a body can be expressed as the energy due to the position of a body;
G.P.E = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now insert the given parameters and solve;
G.P.E = 852 x 9.8 x 3.5 = 29223.6J