Answer:
Explanation:
Expression for times period of a satellite can be given as follows
Time period T = 1.8 x 60 x 60
= 6480
T² =
where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.
6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM
GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²
= 3.96 X 10¹⁴
Expression for acceleration due to gravity
g = GM / R² where R is radius of satellite
20 = 3.96 X 10¹⁴ / R²
R² = 3.96 X 10¹⁴ / 20
= 1.98 x 10¹³ m
R= 4.45 x 10⁶ m
Answer:
The ratio is
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is 
Generally from Kepler's third law

Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that 
=> ![[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%5E2%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E3)
Here
is the period of Deimos
and
is the period of Phobos
So
![[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D)
=> ![\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%7D%7BT_2%7D%20%20%3D%20%20%5B%5Cfrac%7B23500%20%7D%7B9380%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D)
=>
Answer:
<em>F equals 3 N and the object remains stationary</em>. (second option in the list)
Explanation:
For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.
Answer:
42.9 μF
Explanation:
V = 3.50 V, Q = 150 μC
C = Q/V = 150/3.50 μF = 42.9 μF