Thermal energy from the coffee is transferred to the mug.
Explanation:
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Answer:
a) E = 0
b) 
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) 
From which we have;

E = 0/A = 0
E = 0
b) 


By Gauss theorem, we have;

Therefore, we get;

The electrical field outside the spherical shell


Therefore, we have;

Answer:
v = 7.95 m/s
Explanation:
Given that,
Wavelength of a wave, 
Frequency of a wave, f = 15 Hz
We need to find the speed of the wave. The speed of a wave is given by :

So, the wave move with a speed of 7.95 m/s.
Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.