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egoroff_w [7]
3 years ago
10

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

rth with a speed of 0.99542 c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 47.0 km .
Part A As measured by the scientist, how much time does it take the particle to travel the 47.0 km to the surface of the earth?
Physics
1 answer:
monitta3 years ago
7 0

To solve this problem, it is necessary to apply the concepts related to speed of light and the kinematic equations of speed description.

The speed by definition can be expressed as the path of a particle during a certain time, in our case,

v = \frac{d}{t}

Where,

d= distance

v = speed

c = Speed of light (3*10^8m/s)

Replacing to find the time we have,

t=\frac{d}{v}

t = \frac{47000}{0.99542*(c)}

t = \frac{47000}{0.99542*(3*10^8)}

t = 1.57*10^{-4}s

Therefore the time is 1.57*10^{-4}s

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The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever
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Answer:

Approximately 5.19 \times 10^{-5}\; \rm N.

Explanation:

Let G denote the gravitational constant. (G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}.)

Let M and m denote the mass of two objects separated by r.

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}.

In this question: M = 7 \times 10^{22}\; \rm kg is the mass of the moon, while m = 1\; \rm kg is the mass of the water. The two are r = 3\times 10^{5}\; \rm km apart from one another.

Important: convert the unit of r to standard units (meters, not kilometers) to reflect the unit of the gravitational constant G.

\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m.

\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}.

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