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GaryK [48]
2 years ago
13

A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt

er it passes through. One such particle, initially traveling at 2.40 ✕ 106 m/s in a straight line, decreases in speed to 1.56 ✕ 106 m/s over a distance of 1.22 km.
(a) What is the magnitude of the force experienced by the muon?
(b) How does this force compare to the weight of the muon?
|F|/Fg =______
Physics
1 answer:
anyanavicka [17]2 years ago
6 0

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

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A 2000 kg car moves along a horizontal road at speed vo
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Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
3 years ago
A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about 3.0 x 10-5 22 of resi
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Answer:

13.5 x 10^-9 A

Explanation:

Yes

6 0
2 years ago
A paper airplane is thrown westward at a rate of 6 m/s. The wind is blowing at 8 m/s towards the north. Which of the following d
Andrew [12]

since airplane is thrown towards west with speed 6 m/s

while air is blowing with speed 8 m/s towards north

so here the net speed of air plane will be the resultant of airplane speed and wind speed always

SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s

so correct answer must be 1st option


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2 years ago
A train is moving at a speed of 50 km/h. How many hours will it take the train to travel 600 kilometers?
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Answer:

12 hours

Explanation:

600 divided by 50 is 12

6 0
3 years ago
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SVEN [57.7K]

Answer:The human eye is sensitive to yellow-green light having a frequency of about 5.5*10^{14} ... What is the energy in joules of the photons associated with this light? ... As the wavelength and frequency of a wave are related, we can find the energy ... In order to find this value, we need Planck's Constant, h=6.626×10−34 J⋅s h ...

Explanation:

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3 years ago
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