Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
0.66
Explanation:
By using the formula
u = v^2 / r g
Where u is coefficent of friction
u = 23.5 × 23.5 / (85 × 9.8)
u = 0.66
Answer: |F| = 1.28 x 10⁵ N
Explanation:
an impulse results in a change of momentum
FΔt = mΔv
F = m(vf - vi)/t
F = 2000(0 - 32) / 0.5
F = -128,000
|F| = 1.28 x 10⁵ N
In general heating up water Is easier because the water starts out colder, meaning that it will absorb heat easier. Once it is heated there is almost enough heat in there so it starts absorbing it at a slower rate, slowing down the process.
Hi there!
Begin by using Gauss' Law to find the electric field.
E = Electric field (N/C)
dA = differential area element
Q = enclosed charge (C)
ε₀ = Permittivity of free space (8.85 * 10⁻¹² C²/Nm²)
We can construct a large cylinder around the wire in order to determine the electric flux. The electric field lines will pass through the LATERAL surface area of the cylinder, so:
Where 'L' is the length of the cylinder and 'r' is the distance from the capacitor.
The enclosed charge is equivalent to the charge per meter length (λ) multiplied by the length, so:
We can rewrite the dot product as EA (where cosθ = 1 since the normal vector points in the direction of the field).
A = the lateral surface area of a cylinder, so:
Rearrange to solve for 'E'.
a)
Plug in the distance into 'r'.
b)
Repeat:
We can see that the distance from the wire is INVERSELY related to the electric field strength by a power of r⁻¹. The field strength DECREASES as the distance INCREASES.
