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tigry1 [53]
3 years ago
6

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

f 0.6 meters?
A.5.52 × 102 hertz
B.2.56 × 102 hertz
C.459 × 102 hertz
D.675 × 102 hertz
E.990 × 102 hertz
Physics
2 answers:
Olegator [25]3 years ago
8 0

Answer:

Frequency of the tone, f=5.52\times 10^2\ hertz

Explanation:

Given that,

Velocity of sound, v = 331 m/s

Wavelength of tone, \lambda=0.6\ m

We need to find the frequency of the tone. The mathematical formula is as follows

f=\dfrac{v}{\lambda}

f=\dfrac{331\ m/s}{0.6\ m}

f = 551.66 Hertz

or

f = 552 Hertz

f=5.52\times 10^2\ hertz

So, the correct option is (a). Hence, this is the required solution.

Levart [38]3 years ago
3 0
I think c hope it help

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The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
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We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

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From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force  is mathematically given as

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What is the frequency of a wave that has a wavelength of 0.39 m and a speed
gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

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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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