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inessss [21]
3 years ago
6

An open pipe is 1.42 m long.

Physics
1 answer:
Varvara68 [4.7K]3 years ago
8 0

Answer:

f(3) = 362.32 Hz

Explanation:

the formula of frequency of open pipe is

f(n) = (n+1)v/2L

n = 0,1,2,3,... (the order)

0 for the first

1 for the second

v = speed of sound

L = the lenght of pipe

f(3) = (2+1) * 343 / 2 * 1,42

f(3) = 3 * 343 / 2.84

f(3) = 362.324 Hz (or you can write 362.32 Hz

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gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

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v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

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3 years ago
X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{
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<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.

\theta=30\° the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at 30\°:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))   (2)

\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m   (3)

Now, the initial energy E_{o}=400keV=400(10)^{3}eV of the photon is given by:

 E_{o}=\frac{h.c}{\lambda_{o}}    (4)

From this equation (4) we can find the value of \lambda_{o}:

\lambda_{o}=\frac{h.c}{E_{o}}    (5)

\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}    

\lambda_{o}=3.102(10)^{-12}m    (6)

Knowing the value of \Delta \lambda and \lambda_{o}, let's find \lambda':

\Delta \lambda=\lambda' - \lambda_{o}

Then:

\lambda'=\Delta \lambda+\lambda_{o}  (7)

\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m  

\lambda'=3.427(10)^{-12}m  (8)

Knowing the wavelength of the scattered photon \lambda'  , we can find its energy E' :

E'=\frac{h.c}{\lambda'}    (9)

E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}    

E'=362.063keV    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron E_{e}, we have to calculate all the energy lost by the photon, which is:

E_{e}=E_{o}-E'  (11)

E_{e}=400keV-362.063keV  

Finally we obtain the energy of the recoiling electron:

E_{e}=37.937keV  

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