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3 years ago

7

A race car travels a circular track at an average rate of 135 mi/hr. The radius of the track is 0.450 miles. What is the centrip

etal acceleration of the car?

2 answers:

kenny6666 [7]3 years ago

7 0

135 * 135 / 0.450 = 40,500 mi/hr2

PilotLPTM [1.2K]3 years ago

4 0

Answer : $a_c=40500\ mi/h^2$ .

Explanation :

It is given that,

Velocity of a race car, v = 135 mi/hr

Radius of the track, r = 0.45 miles

We know that the centripetal acceleration is defined as :

$a_c=\dfrac{v^2}{r}$

r is the radius of the circle.

$a_c=\dfrac{(135\ mi/hr)^2}{0.45\ mi}$

$a_c=40500\ mi/h^2$

So, the centripetal acceleration of the car is $a_c=40500\ mi/h^2$ .

Hence, this is the required solution.

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3 years ago

sergiy2304 [10]

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3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

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Answer:

<h3>a.</h3>

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<h3>b.</h3>

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<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

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$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

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<h2>b</h2>

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$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

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$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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3 years ago