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3 years ago

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A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relati

ve to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

Answer:0.3568 m/s

Explanation:

Given

Velocity of a canoe is 0.4 m/s southeast relative to earth

Representing in vector form

$V_{c}=\left ( 0.4cos45\right )\hat{i}-\left ( 0.4sin45\right )\hat{j}$

Also Velocity of river $\left ( V_r\right )=0.5\hat{i}$

thus velocity of canoe relative to the river

$V_{cr}=V_c-V_r$

$V_{cr}=\frac{0.4}{\sqrt{2}}\hat{i}-\frac{0.4}{\sqrt{2}}\hat{j}-0.5\hat{i}$

$V_{cr}=\left ( \frac{0.4\sqrt{2}-1}{2}\right )\hat{i}-\left ( \frac{0.4\sqrt{2}}{2}\right )\hat{j}$

magnitude of Velocity $=\sqrt{\left ( \frac{0.4\sqrt{2}-1}{2}\right )^2+\left ( \frac{0.4\sqrt{2}}{2}\right )}$

$V_{net}=0.3568 m/s$

$tan\theta =\frac{\frac{0.4\sqrt{2}}{2}}{\frac{0.4\sqrt{2}-1}{2}}$

$\theta =52.47 ^{\circ}$ with east direction in clockwise direction

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

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

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So then if we replace we got:

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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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