What energy transformation takes place in the solar panel?
1 answer:
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Answer:
It increases proportionally
Explanation:
The gravitational force between the Earth and an object on its surface is given by
where
G is the gravitational constant
M is the Earth's mass
m is the mass of the object
R is the Earth's radius
In this problem, the Earth's mass is increased, while the diameter (and therefore, the radius) doesn't change. From the equation, we see that the gravitational force is directly proportional to the Earth's mass: therefore, if the mass is increased, the force will increase as well by the same proportion (for example, if the mass is doubled, the force will double as well)
I have attached the circuit image missing in the question.
Answer:
A) The range of vo is; -6.6V≤ vo ≤-1V
B) σ = 0.1861
Explanation:
A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.
Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0
So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0
Simplifying further; -25 vg - vΔ = 0
From the question, vg = 40mV = 0.04 V
So - 25(0.04) = vΔ
So: vΔ = - 1 V
Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0
So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0
So RΔ = 100kΩ or 100,000Ω from the question.
So, substituting for RΔ, we get,
[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0
Let's put the value of - 1 for vΔ as gotten before.
So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0
Now let's make vo the subject of the equation to get;
-1 - vo = (1 - σ)[2 + (1/σ)]
-1 - vo = 2 - 2σ + (1/σ) - 1
-vo = 1 + 2 - 2σ + (1/σ) - 1
-vo = 2 - 2σ + (1/σ)
vo = - 1 (2 - 2σ + (1/σ))
When σ = 0.2; vo = - 1(2 - 0.4 + 5) =
- 1 x 6.6 = - 6.6V
Also when σ = 1;
vo = - 1(2 - 2 + 1) = - 1V
Therefore, the range of vo is;
- 6.6V ≤ vo ≤ - 1V
B) it will saturate at vo = - 7V
So, from;
vo = - 1 (2 - 2σ + (1/σ))
-7 = - 1 (2 - 2σ + (1/σ))
Divide both sides by (-1)
7 = (2 - 2σ + (1/σ))
Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)
Multiply each term by α to get;
5σ = - 2σ^(2) + 1
So 2σ^(2) + 5σ - 1 = 0
Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861
