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maw [93]
2 years ago
14

Please help me answer the following question!

Physics
1 answer:
Troyanec [42]2 years ago
3 0

Answer:

P = 0

Explanation:

Momentum is defined as the quentity of motion contained in a body. Mathematically, it can be defined as the product of mass and velocity. So, in order to determine the Volkswagen Beetle, at rest, we can use the simple formula, as follows:

where,

P = Momentum of Volkswagen Beetle = ?

m = Mass of Volkswagen Beetle = 750 kg

v = Velocity of Volkswagen Beetle = 0 m/s (since, it is at rest)

Therefore,

P =

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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
Paha777 [63]

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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I think the answer is ruthorford
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A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width
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Answer:

4.4 cm

Explanation:

Given:

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We have:

2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}

Calculating further for the width of the central bright fringe, we have:

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