Answer with Explanation:
We are given that


Charge on proton,q=
a.We have to find the electric potential of the proton at the position of the electron.
We know that the electric potential

Where 


B.Potential energy of electron,U=
Where
Charge on electron
=Charge on proton
Using the formula


Inversely proportional to its frequency. If electromagnetic radiation A has a lower frequency than electromagnetic B, then compared to B, the wavelength of A is...? - equal - shorter - longer - exactly half the length of
This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.
Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency
fᵇ = |f₁ - f₂|
substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz
The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
</span>
Therefore, the answer is C. 4
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
Answer:
Final Velocity = √(eV/m)
Explanation:
The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference
W = (2e) × V = 2eV
And this work is equal to change in kinetic energy
W = Δ(kinetic energy) = ΔK.E
But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0
ΔK.E = ½ × (mass) × (final velocity)²
(Velocity)² = (2×ΔK.E)/(mass)
Velocity = √[(2×ΔK.E)/(mass)]
ΔK.E = W = 2eV
mass = 4m
Final Velocity = √[(2×W)/(4m)]
Final Velocity = √[(2×2eV)/4m]
Final Velocity = √(4eV/4m)
Final Velocity = √(eV/m)
Hope this Helps!!!