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3 years ago

A fisherman sees 7 wave crests go by in 10.0 s. The crests are 2.43 m apart. Find the period of the wave. (Unit = s)

Physics

2 answers:

Gekata [30.6K]3 years ago

8 0

Velocity of the wave is 1.7 m/s

Explanation:

As we know that the frequency of the wave is defined as the number of waves crossing a fixed point per second

So here we will have

Also we know that the distance between two consecutive crest or two consecutive trough is known as wavelength

So here the wavelength of the wave is given as

now we have

nasty-shy [4]3 years ago

5 0

Answer:

Explanation:

Wavelength=2.43

frequency=0.700 hz

t=1.43

v=1.70

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Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

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0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

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Answer:

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

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:

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

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

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:

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

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

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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A fisherman sees 7 wave crests go by in 10.0 s. The crests are 2.43 m apart. Find the period of the wave. (Unit = s)​

A fisherman sees 7 wave crests go by in 10.0 s. The crests are 2.43 m apart. Find the period of the wave. (Unit = s)