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


Answer:
The fraction of mass that was thrown out is calculated by the following Formula:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
We know that Force on a moving object is equal to the product of its mass and acceleration given as:
F = ma
And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²
Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.
Case 1:
Hot balloon of mass = M
acceleration = a
Upward force due to hot air = F = constant
Gravitational force downwards = Mg
Net force on balloon is given as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (balloon is moving downwards so Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:
Net Force is given as:
-m(a/2) = mg - F (Balloon is moving upwards so F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Calculating the fraction of the initial mass dropped:
![M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]](https://tex.z-dn.net/?f=M-m%20%3D%20%5Cfrac%7BF%7D%7Bg-a%7D%20-%20%5Cfrac%7BF%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B1%7D%7Bg-a%7D%20-%20%5Cfrac%7B1%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%28g%2B%28a%2F2%29%29%20-%20%28g-a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7Bg%2B%28a%2F2%29%20-%20g%20%2B%20a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%283a%2F2%29%7D%7Bg%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7Bag%7D%7B2%7D%7D%20%5D)
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

Answer:
A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.
Explanation:
This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂
Since F₁ = 20 N, then F₂ = -F₁ = -20 N
So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.
So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.