Explanation:
1. Mass of the proton, 
Wavelength, 
We need to find the potential difference. The relationship between potential difference and wavelength is given by :



V = 45.83 volts
2. Mass of the electron, 
Wavelength, 
We need to find the potential difference. The relationship between potential difference and wavelength is given by :




V = 84109.27 volt
Hence, this is the required solution.
Let the sphere is having charge Q and radius R
Now if the proton is released from rest
By energy conservation we can say



now take square root of both sides

so the proton will move by above speed and
here Q = charge on the sphere
R = radius of sphere

Answer:
Heat
Friction is what causes heat.
Brainliest always helps.
<span>It’s 2/5 MR^2 where M is mass and R is the radius of the bas</span>