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3 years ago

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A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

and slows at 7.0 m/s2. How far does the car go before it stops?

1 answer:

mr Goodwill [35]3 years ago

3 0

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

$d_1 = v* t$

$d_1 = 20 * 0.8$

$d_1 = 16 m$

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2*(-7)*d_2$

$d_2 = 28.6 m$

so the total distance covered by it

$d = d_1 + d_2$

$d = 16 + 28.6 = 44.6 m$

<em>so it will cover a total distance of 44.6 m</em>

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Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

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3 years ago

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