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3 years ago

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An ice skater accelerates backwards for 5.0 seconds to a final speed of 12 m/s. If the acceleration backwards was at a rate of 1

.5 m/s^2 what was the skaters initial speed?

1 answer:

Lostsunrise [7]3 years ago

4 0

Answer:

Vo = 4.5 [m/s]

Explanation:

In order to solve this problem, we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 12 [m/s]

Vo = initial velocity [m/s]

a = acceleration = 1.5 [m/s²]

t = time = 5 [s]

Now replacing:

$12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]$

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Answer:

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

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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:

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

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

$c = 3x10^8 m/s$

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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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