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1 year ago

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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?

1 answer:

Paha777 [63]1 year 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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2 years ago

In a race, you run 3000 meters east in 21 minutes. What is your velocity in km/min?

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Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

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Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

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So for satellite X, the orbital period $T_{X}$ is:

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Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

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2 years ago

Read 2 more answers

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Answer:

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2 years ago

proton is accelerated to a speed of 0.93c, what is the 1. A proton has a mass of 1.673 x 10-27 kg. If the proton is accelerated

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Answer : The correct option is, (B) $1.3\times 10^{-18}kg.m/s$

Explanation : Given,

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Formula used for relativistic momentum of the proton is:

$p=\frac{m_ov}{\sqrt{1-\frac{v^2}{c^2}}}$

where,

p = relativistic momentum of the proton

$m_o$ = mass of proton

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Now put all the given values in the above formula, we get:

$p=\frac{(1.673\times 10^{-27}kg)\times (0.93c)}{\sqrt{1-\frac{(0.93c)^2}{c^2}}}$

$p=\frac{(1.55589\times 10^{-27}c)}{0.367}$

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