Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Answer:
F = 1263.03 N
Explanation:s
given,
mass of the disk thrower = 100 Kg
mass of the disk = 2 Kg
angular speed of the disk = 4 rev/s
arm outstretched = 1 m
centripetal force of the disk in the circular path
F = m ω² r
ω = 4 x 2 x π
ω = 25.13 rad/s
F = m ω² r
F = 2 x 25.13² x 1
F = 1263.03 N
hence, centripetal force equal to the F = 1263.03 N
Answer:
V₀ = 5.47 m/s
Explanation:
The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 3.04 m
θ = Launch Angle = 41.7°
V₀ = Minimum Launch Speed = ?
g = 9.81 m/s²
Therefore,
3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)
V₀² = 3.04 m/(0.10126 s²/m)
V₀ = √30.02 m²/s²
<u>V₀ = 5.47 m/s</u>
Answer: I have no idea either i need help aswell
Explanation:
