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Degger [83]
2 years ago
9

A fighter plane is flying overhead at mach 1.20. What angle does the wave front of the shock wave produced make relative to the

plane's direction of motion (in degrees)?
Physics
1 answer:
Sever21 [200]2 years ago
7 0

Answer: 56.44°

Explanation:

<u>Given:</u>

  • Let u represent the current speed of the plane, <u>1.2 Mach</u>

<em>Converting to SI Units (m/s):</em>

= (1.2 mach)(340 ms^-1 / 1 Mach)

u  = 408 m/s

  • Speed of sound in air, v = 340 m/s

<u>Find:</u>

  • Angle the wave front of the shock wave relative to the plane's direction of motion, θ

We have, sinθ = speed of sound / speed of object

               sinθ = v / u

                   θ = sin^-1 (v / u)  

                      = sin^-1 (340 / 408)

                   θ = 56.44°

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svetoff [14.1K]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

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3 years ago
A cyclist is moving at a speed of 15 m/s. If the combined mass of the bike and person is 100 kg,
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Answer:

Explanation:

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2 years ago
Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m
Leno4ka [110]

Answer: 1.593*10^{17} photons released if the laser is used 0.056 s during the surgery

                           

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

                            E =\frac{hc}{\lambda}

Where \lambda is the wavelength, h is the Planck's constant and  h is the speed of light

              h = 6.626*10^{-34} \frac{m^{2} kg }{s}  -> Planck's constant

              c = 3*10^{8} \frac{m}{s}  -> Speed of light

So, replacing in the equation:

                E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}

Then, the energy of each released photon by the laser is:

                E = 3.867*10^{-19} \frac{J}{photons}

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

               \frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

              2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}

And by doing a simple rule of three, if 2.844*10^{18} photons are released every second, then in 0.056 s:

            0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons are released during the surgery

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Soloha48 [4]
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