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3 years ago

9

A 1400 kg car is moving at 12 m/s when the driver stops a car what increase in the thermal energy of the car and it surroundings

results from a find the brakes to bring the car to a stop

2 answers:

butalik [34]3 years ago

5 0

KE = 1/2 mv^2

in this case, the initial kinetic energy which is converted to heat is

KE = 1/2 1400 (12)^2

KE = 100,800 J

seropon [69]3 years ago

3 0

KE = 1/2 1400 (12)^2

KE = 100,800 J

Answer: 1.0 x 10^5 J

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<h3>a.</h3>

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For our problem, after travelling 1 cm:

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

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

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

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After travelling 2 cm:

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$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

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

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

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

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

$od( 1\ cm) = 0.2606$

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

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

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$od( 2\ cm) = 0.5211$

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