By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m
<h3>
What is the orbital period?</h3>
This would be the time that a given body does a complete revolution in its orbit.
It can be written as:
Where:
- π = 3.14
- G is the gravitational constant = 6.67*10^(-11) m^3/(kg*s^2)
- M is the mass of the sun = 1.989*10^30 kg
- r is the radius, which we want to find.
Rewriting the equation for the radius we get:
![T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }](https://tex.z-dn.net/?f=T%20%3D%20%5Csqrt%7B%5Cfrac%7B4%2A%5Cpi%20%5E2%2Ar%5E3%7D%7BG%2AM%7D%20%7D%5C%5C%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7BT%5E2%2AG%2AM%7D%7B4%2A%5Cpi%20%5E2%7D%20%7D)
Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s
Replacing the values in the equation we get:
![r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B%282.3655%2A10%5E8%20s%29%5E2%2A%286.67%2A10%5E%7B-11%7D%20m%5E3%2F%28kg%2As%5E2%29%29%2A%281.989%2A10%5E%7B30%7D%20kg%29%7D%7B4%2A3.14%20%5E2%7D%20%7D%20%3D%204.29%2A10%5E%7B11%20%7Dm)
So the orbital radius is 4.29*10^11 m
If you want to learn more about orbits, you can read:
brainly.com/question/11996385
Answer:
Temperature is also a condition that affects the speed of sound. Heat, like sound, is a form of kinetic energy. Molecules at higher temperatures have more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly.
Answer:
Recoil speed, 
Explanation:
Given that,
Mass of the comet fragment, 
Speed of the comet fragment, 
Mass of Callisto, 
The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,
V is recoil speed of Callisto immediately after the collision.
So, the recoil speed of Callisto immediately after the collision is
Answer:
Approximately 
and approximately 
.
Explanation:
Let 
and 
denote the capacitance of these two capacitors.
When these two capacitors are connected in parallel, the combined capacitance will be the sum of and . (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)
.
On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:
.
(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field (
) between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored 
stays the same.)
The question states that:
, and
.
Let the capacitance of these two capacitors be 
and 
. The two equations will become:
.
From the first equation:
.
Hence, the 
in the second equation here can be replaced with 
. That equation would then become:
.
Solve for 
:
.
.
.
Solve this quadratic equation for :
or 
.
Substitute back into the equation for :
In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately 
and approximately (both values are rounded to two significant digits.)
