The amount of energy for a coulomb of charge at one end of a conductor is 40 joules. The amount of energy for a coulomb of charge at the other end of the conductor is 20 joules. The potential difference across the ends of the conductor is ______.

A.) 60 joules per coulomb
B.) 60 joules
C.) 20 joules per coulomb
D.) 20 joules

Answers

Answer 1
Report this clown who put the first answer he’s trying to get your ip .

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Identify and describe one past geologic process that formed copper ore deposits.

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tornado and land

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Which of the following does NOT create a mechanical wave? Text to speech
A. a shaking rope
B. a guitar being played
C. dropping a rock in a pond
D. light coming through a window

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light coming through a window!!

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light coming through a window

HELP ME PLEASEEEEEEEEEEEEEE

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It makes energy for the cell

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probably won't help now but the answer is the 4th one lol

Explanation:

Suppose that you make a series RC circuit with a capacitor and a known resistor that has a 5% tolerance: R= 5.20 ± 0.26kΩ. You plan is to measure the time constant of the circuit in order to determine the value of the capacitor. You do ten trials of your experiment and measure the time constant each time, and then you calculate the average and standard error of the ten results to determine that τ= 2.150 ± 0.002s.

Which of the following is the best estimate of the uncertainty on your determination of the value of your capacitor?
a. 1%
b. 0.1%
c. 5%

Answers

Answer:

correct answer is C

Explanation:

The time constant of an RC circuit is

           τ = RC

so to find the capacitance

          C = τ/ R

          C = 2.150 / 5.20 10³

          C = 4.13 10⁻⁴ F

to find the error we use the worst case

         ΔC = | [tex]|\frac{dC}{d \tau }| \ \Delta \tau + | \frac{dC}{dR} | \ \Delta R[/tex]

the absolute value guarantees that we find the worst case, we evaluate the derivatives

          ΔC = 1 /R Δτ + τ/R²  ΔR

the absolute values ​​of the errors are

          Δτ = 0.002 s

          ΔR = 0.3 kΩ

we substitute

           ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)²   0.3 10³

           ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵

           ΔC = 1.77 10⁻⁵ F

the uncertainty or error must be expressed with a significant figure

            ΔC = 2 10⁻⁵ F

the percentage error is

            Er% =[tex]\frac{\Delta C}{C} \ 100[/tex]

            Er% = [tex]\frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100[/tex]

            Er% = 4.8%

the correct answer is C

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