Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
Answer:
I think the answer is b am sorry if it is wrong
Explanation:
Answer:
The motion is over-damped when λ^2 - w^2 > 0 or when 
> 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86
Explanation:
Using the newton second law
k is the spring constante
b positive damping constant
m mass attached
x(t) is the displacement from the equilibrium position
Converting units of weights in units of mass (equation of motion)
From hook's law we can calculate the spring constant k
If we put m and k into the DE, we get
Denoting the constants
2λ = 
= 
λ = b/0.215
λ^2 - w^2 = 
This way,
The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86
The answer is C , you’re welcome
<span>Distance in Miles (.25) x 3600 (seconds in an hour) / time in seconds = 200 MPH. Drag racing calculates the top speed via a speed trap starting 66 feet from the finish line.</span>
