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SSSSS [86.1K]
3 years ago
6

The table below shows the average distances of Venus and Earth from the sun. -Name of Planet -Distance from the sun (in AU) Venu

s 0.72 Earth 1.0 What is the difference between the orbital periods of Venus and Earth? a. 1.72 years b. 1.62 years c. 0.62 years d. 0.28 years
Physics
2 answers:
Serhud [2]3 years ago
6 0
<span>Venus goes around faster than Earth, so its period must be less than Earths 1.0 years. That eliminates the first two answers. My recollection is that T^2 = k r^3, relating period of orbit to radius, so T should be (0.72/1.00)^3/2 = 0.61 of the Earths period, so 0.62 is ans. </span>
Paul [167]3 years ago
6 0

Answer: 0.39 years (Venus orbits the Sun in 0.62 Earth years)

Explanation:

The orbital period is related to the distance of the planet from the star as:

T² = R³

where T is in years and R is in AU.

<u>Orbital period of Venus:</u>

T² = (0.72)³

⇒ T = 0.62 years

<u>Orbital period of Earth </u>

T² = (1.0)³

⇒T = 1 year

The difference between the orbital periods of Venus and Earth is: (1 -0.62) = 0.38 years. Venus orbits the Sun in 0.62 Earth years.

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Explanation:

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You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

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The equation can be rewritten as

a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]

where L is the length of one car.

The same equation can be written considering the first 7 cars:

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where

7L is the distance covered by the 7 cars

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We still have

u = 0

And the acceleration is constant so it is

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Substituting into the equation, we can find t':

7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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