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Anika [276]
2 years ago
8

A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

is the ball moving?
Physics
1 answer:
Paha777 [63]2 years ago
3 0

The speed of the ball moving is

v = 3.94 \times 10 {}^{ - 25}m/s

what is momentum?

The momentum p of a classical object of mass m and velocity v is given by pclassical =mv.

For photons with wavelength λ,this equation does not hold.Instead, the momentum of the Photon is given by p Photon = h/λ

where,h is the planck's constant.

The momentum of the red Photon is

given:

h = 6.626 \times 10 {}^{ - 34}kgm {}^{2}/s(plancks \: constant)

λ = 700 \times 10 {}^{ - 9} m(photons \: wavelength)

p \: photons =  \frac{6.626 \times 10 {}^{ - 34}kgm {}  ^{2}/s }{700 \times 10 {}^{ - 9}m  }

= 9.47 \times 10 {}^{ - 28}kgm /s

since,the Photon and the ping-pong ball have the same momentum,we have

pball = pphotons =  \frac{6.626 \times 10 {}^{ - 34}kgm/s  }{700 \times 10 {}^{ - 9} }

pball = mv,m = 2.40 \times 10 {}^{ - 3}kg

v = 3.94 \times 10 {}^{ - 25} m/s

Therefore, if the red photon and the ping-pong ball have the same momentum, the ping-pong ball must have a speed of approximately

v = 3.94 \times 10 {}^{ - 25}m/s

learn more about momentum of photon from here: brainly.com/question/28197406

#SPJ4

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In the first part of the problem, light moves from glass to air (n_a=1.00) and the critical angle is \theta_c = 30.8^{\circ}. This means that we can find the refractive index of glass by re-arranging the previous formula:
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\theta_c = \arcsin ( \frac{n_2}{n_1} )=\arcsin( \frac{n_w}{n_g} )=\arcsin( \frac{1.33}{1.95} )=\arcsin(0.68)=43.0^{\circ}
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Explanation:

a) In the center of the slab there are no charges inside, so by Gauss's law the electric field is zero,

Another way of analyzing it is that the charge on one side of the crockery creates an outgoing electric field in the center, the charge on the other side of the crockery creates a field of equal magnitude, but in the opposite direction, so the resulting field is zero. .

b) Let's use Gauss's law to calculate the electric field, let's use as cylinder a Gaussian surface with the base parallel to the faience, so the scalar product is reduced to the algebraic product

         Ф = ∫ E. dA = qint /ε₀

 

The slab area is A

let's use the concept of charge density

         ρ = qint / V

the volume of the slab is the area times the thickness

         V = A x

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       2E A = ρ A x /ε₀  

          E = ρ x / 2ε₀  

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for this case

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Therefore, option "D" is the correct answer

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