Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are se
1 answer:
Answer:
The force between them when they are separated by 12 cm is 5 N.
Explanation:
Distance between two point charges, d = 6 cm
The attractive force between them is 20 N, F = 20 N
Let F' is the force between them when they are separated by 12 cm, d' = 12 cm
The force between point charges is given by the formula as :
........(1)
New force,
............(2)
From (1) and (2) :
So, the force between them when they are separated by 12 cm is 5 N. Hence, this is the required solution.
You might be interested in
Answer:
1. <u>Potential energy</u>, 2. <u>Potential and kinetic energy</u>, 3. <u>Potential and kinetic energy</u>, 4. <u>Potential and kinetic energy</u>, 5. <u>Potential energy</u>
Explanation:
We note that the total mechanical energy (M.E.) of the body is given as follows;
M.E. = K.E. + P.E. = Constant
Where;
K.E. = The kinetic energy of the body = (1/2)·m·v²
P.E. = The potential energy of the body = m·g·h
m = The mass of the person
v = The velocity with which the person is in motion
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height of the person above the ground
The length of the rope = 20 m
The initial height at location 1, h₁ = 40.0 m
At location 1, the velocity, v₁ = 0.00 m/s
The mechanical energy, M.E. = K.E.₁ + P.E.₁
∴ K.E.₁ = 0 and P.E.₁ = m ×9.81×40
M.E. = (1/2) ×m ×0² + m ×9.81×40
∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy
At location 2, the velocity, v₂ = 10.0 m/s
The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40
∴ K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m
M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.
At location 3, the velocity, v₃ = 20.0 m/s
The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20
∴ K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m
M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.
At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m
The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15
∴ K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m
M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.
At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m
The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10
∴ K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m
M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.
Therefore, we have;