B) the current will decrease
The force on charge Y is the same as the force on charge X
Explanation:
We can answer this problem by applying Newton's third law of motion, which states that:
"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by
(1)
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.
Learn more about Newton's third law:
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Answer:
(B) cash inflows are moved earlier in time.
Explanation:
The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form
The formula to compute the payback period is shown below:
Payback period = Initial investment ÷ Net cash flow
where,
The net cash flow = annual net operating income + depreciation expenses
The payback period of the project decreases when the accumulated starting year cash flows increases that results the movement of the cash inflows earlier in time
I think Im gonna have to go with C 6.00 T/s but Im not sure
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
