What is one way to lower gravitational potential energy?
1 answer:
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Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
