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Lady_Fox [76]
2 years ago
11

In an electrical circuit, what happens to the current flowing through the wire if the initial voltage of 18 V is doubled, and th

e initial resistance of 35 Ohms is reduced by a factor of 4?
ANSWER QUICKLY PLEASE! ITS URGENT!
Physics
2 answers:
Evgen [1.6K]2 years ago
5 0

Answer:

I hope this helps you :)

Explanation:

The current is directly proportional to the voltage and inversely proportional to the resistance. ... So doubling or tripling the resistance will cause the current to be one-half or one-third of the original value.

Genrish500 [490]2 years ago
3 0

The current is increased by a factor of 8.

From 18/35 A. to 144/35 A.

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The found acceleration in terms of h and t is:

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(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

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<h3>Stage 2</h3>

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<h3>y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\</h3><h3 /><h3>Stage 3</h3>

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y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2

<h3>Total Height</h3>

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<h3 /><h3>Acceleration</h3>

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