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faltersainse [42]
2 years ago
12

Pls help in these 2 questions

Physics
1 answer:
Nadya [2.5K]2 years ago
4 0

Answer:

Taking forces along the plane

F cos θ - M g sin θ -100 = M a       net of forces along the plane

F = (M a + M g * .5 + 100) / .866     solving for F

F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N

F = 707 N acting along the plane

Fn = F sin θ + M g cos θ       forces acting perpendicular to plane

Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons   forces normal to plane

(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)

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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

3 0
2 years ago
9. What is the gravitational potential energy of a 61.2 kg person standing on the roof of a 10-storey building? (Each storey is
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Answer:

15009

Explanation:

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PE = 15009.3

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The moment of water from ocean through the atmosphere and back
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There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

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Answer:

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