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


The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:

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Answer:
60*12.0= 720 = v/60 * 12.0 squared which is 1,728
Explanation:
Horizontal velocity component: Vx = V * cos(α)
Answer:
F = 6.27 x 10 ¹⁹ N
Explanation:
Given
m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol
q₁ = % * [m * N * A * e / M ]
q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]
q₁ = 3.54 x 10⁶ C
q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]
q₂ = 1.773 x 10⁶ C
Now to determine the electrostatic force con use the equation
F = K * q₁ * q₂ / d²
K = 8.99 x 10 ⁹
F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²
F = 6.27 x 10 ¹⁹ N
Answer: A, C, E
Explanation:
Gamma rays, Microwaves, and Radio waves