Given the distance r = 2/1000 m, the force between them F =
0.0104 N, the mass of the two object can be calculated using formula:
F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2
And where G = is the gravitational constant (6.67E-11 m3 s-2
kg-1)
The mass of the two objects are 24.96 kg
Answer: Sean is standing still, and Rhea is running toward Sean while kicking the ball
Explanation: Your welcome :)
D:the electrons from being attracted to the grid instead of the anode
We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N