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)
=>
Answer:
The average speed of the earth in its orbit is 
Explanation:
The average distance between the Earth and the Sun is 
.
The average speed of the earth in its orbit can be found by the next equation :
(1)
Where r is the radius and T is the period.
In this case, the orbit of the Earth can be considered as a circle
(
) instead of an ellipse.
It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.
Notice that to express the period in terms of seconds, the following is needed:
⇒ 
Then, equation 1 can be used:
Answer:
Explanation:
In first case we are interested in one time 6 in six rolls
Thus probability = number of chances required/Total chances
= 1/6
Similarly in the second case probability = 2/12 = 1/6
In the same way in last case probability = 100/600 = 1/6
The probability is the same . Thus all the cases has equal chances
Answer: 8.6 µm
Explanation:
At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:
Ey =Emax cos (kx-ωt)
So, we can write the following equality:
ω= 2.2 1014 rad/sec
The angular frequency and the linear frequency are related as follows:
f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec
In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.
The wavelength, speed and frequency, are related by this equation:
λ = c/f
λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.
Answer:
a = 3.33 m/s²
Explanation:
The horizontal acceleration of the dog can be found by using Newton's Second Law of Motion as follows:
F = ma
where,
F = Unbalanced force applied on the dog = 11 N - 10 N = 1 N
m = mass of the dog = 0.3 kg
a = horizontal acceleration of dog = ?
Therefore,
1 N = 0.3 kg(a)
a = 1 N/0.3 kg
<u>a = 3.33 m/s²</u>
