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artcher [175]
4 years ago
6

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb

it, assuming a circular orbit?
Physics
1 answer:
Maslowich4 years ago
5 0

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

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kondor19780726 [428]

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3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

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7 0
4 years ago
When Maggie applies the brakes of her car, the car slows uniformly from 15.3 m/s to 0 m/s in 2.23 s. How far ahead of a stop sig
nirvana33 [79]

Answer:

The answer is 34.119m

Explanation:

You need to pay attention to the word "uniformly". It means there is no acceleration thus the physics of this problem respond to the uniform rectilinear motion equations:

  • Xf = Xo + vt
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where:

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Xf = 34.119m

3 0
4 years ago
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