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denpristay [2]
3 years ago
9

an athlete runs 5.4 laps around a circular track that is 400.0 m long. If this takes 540 s, what is the average velocity of the

athlete? What is the average speed of the athlete over the complete distance of the run?
Physics
1 answer:
Slav-nsk [51]3 years ago
6 0

Answer:

Average velocity is 0.296 m/s.

Average speed is 4.0  m/s.

Explanation:

Given:

Distance of the circular track is, D=400.0\ m

Number of laps ran is, n=5.4

Time taken for the run is, t=540\ s

Now, total distance covered in 5.4 laps = D_T=D\times n=400\times 5.4=2160\ m

Also, since the path is a circle, the final position of the athlete after 5.4 laps will be 0.4 of 400 m ahead of the starting point.

Distance covered in 0.4 laps is, \textrm{Displacemet}=0.4\times 400=160\ m

Therefore, the displacement of the athlete will be 160 m as the athlete is 160 m ahead of the starting point and displacement depends on the initial and final points only.

Now, average velocity is given as:

v_{avg}=\frac{\textrm{Displacemet}}{t}=\frac{160}{540}=0.296\ m/s

Average speed is the ratio of total distance covered to total time taken.

So, average speed = \frac{D_T}{t}=\frac{2160}{540}=4\ m/s

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Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

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b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

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