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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# If you wish to take a picture of a bullet traveling at 500 m/s , then a very brief flash of light produced by an discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one RC constant is acceptable, and given that the flash is driven by a 600-mF capacitor, what is the resistance in the flash tube?

The resistance in the flash tube is $3.33×{10}^{-3}\mathrm{\Omega }$

See the step by step solution

## Step 1: Definition of RC constant and resistance.

RC constant: The RC time constant is a measurement that tells us how long a cap will take to charge to a specific voltage level.

Resistance: The term "resistance" refers to anything that stands in the way of current flow.

## Step 2: Given information

Bullet Speed= 500 m/s

Capacitor= $600\mathrm{\mu F}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{\mu F}}\right)=6.00×{10}^{-4}\mathrm{F}$

## Step 3: Calculating the time taken by a bullet

Let us solve the given problem.

In this problem, we calculate the resistance necessary in and RC circuit to take a picture of a bullet with minimal blurring. The bullet travels at a speed of $\mathrm{V}=500\mathrm{m}/\mathrm{s}$. If $\mathrm{x}=1\mathrm{mm}\left(\frac{{10}^{-3}}{1\mathrm{mm}}\right)=1\text{'}{10}^{-3}\mathrm{m}$ of motion is allowed during one time constant is acceptable, we first calculate the time it takes the bullet to travel 1 mm. This gives

$\mathrm{t}=\frac{\mathrm{x}}{\mathrm{v}}\phantom{\rule{0ex}{0ex}}=\frac{1\text{'}{10}^{-3}\mathrm{m}}{500\mathrm{m}/\mathrm{s}}\phantom{\rule{0ex}{0ex}}=2.00\text{'}{10}^{-6}\mathrm{s}$

## Step 4: Calculating the resistance

The time has to be equal to the time constant of the circuit.

For the capacitance,$\mathrm{C}=6.00\text{'}{10}^{-4}\mathrm{F}$ , we have

role="math" localid="1656399114049" $\mathrm{t}=\mathrm{RC}\phantom{\rule{0ex}{0ex}}\mathrm{R}=\frac{\mathrm{t}}{\mathrm{C}}\phantom{\rule{0ex}{0ex}}\mathrm{R}=\frac{2×{10}^{-6}\mathrm{s}}{6.00×{10}^{-4}\mathrm{F}}\phantom{\rule{0ex}{0ex}}\mathrm{R}=3.33×{10}^{-3}\mathrm{\Omega }$

Therefore, the resistance is $3.33×{10}^{-3}\mathrm{\Omega }$ .