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Mumz [18]
3 years ago
8

How is the voltage drop ΔV across the resistor related to the current I and the resistance R of the resistor? (Use any variable

or symbol stated above as necessary.)
Physics
1 answer:
MissTica3 years ago
7 0

P = ΔV²/R

P = I²R

<u>Explanation:</u>

As a q passes by a resistor, it expends a (q V) where V is the potential drop between the resistor. This energy goes into radiation, much like the type a glob of putty that drops off a rock transforms its potential energy to heat when it strikes the bottom.

The transformation of potential energy into heat is referred to as dissipation. The power consumed in a resistor is the energy consumed per time. If an amount of Δq moves through the resistor in a time Δt, the power P = ΔqV/ Δt

                         P = ΔV²/R

                         P = I²R or IV

Where I is the current through the resistor, R and V is the voltage drop crossed it. The formula P = IV also provides the energy produced by a battery if I is the current spreading from the battery and V is its voltage.

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adell [148]

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Explanation:

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lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

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1. F

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So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

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So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

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3. 6F

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So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

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8 0
2 years ago
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A well is being dug. A 4.5-kg bucket is filled with 28.0 kg of dirt and pulled vertically upward at a constant speed through a d
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Using the formula:

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Therefore, the work done is 2, 925 Joules

Learn more about work done here:

brainly.com/question/25573309

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meriva

Answer:

true

Explanation:

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