Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is 
Explanation:
From the question we are told that
The semi - major axis of the rocky debris 
The semi - major axis of Planet D is 
The orbital period of planet D is 
Generally from Kepler third law
Here T is the orbital period while a is the semi major axis
So
=> ![T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%20T_D%20%2A%20%20%5B%5Cfrac%7Ba_R%7D%7Ba_D%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=> ![T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=>
Yes that looks correct to me. good luck!!
Answer:heat brings it up then down
Explanation:
Hello,
<u>Solution for A:</u>
Force = 3.00N
Mass = 0.50 Kgs
Time = 1.50 Seconds
According to newton's second law of motion;
Force = Mass times Acceleration(a)
3.00 = 0.50 * a
a = 3.00/0.50 = 6.00 m/s^2
We know that acceleration = Velocity / time
So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2
<u>Solution for B:</u>
The net force = 4.00N - 3.00N = 1.00N to the left
Force = 1.00N
Mass = 0.50Kg
Time = 3.00 Seconds
Again; F = MA (Where F is force, M is mass and A is acceleration)
1.00N = 0.5 * A
A = 1/0.5 = 2 m/s^2
Velocity = Acceleration * Time = 2 * 3 = 6 m/s
I'm pretty sure this is A
