Answer:
mass of the planet: 
Explanation:
When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.
We know the orbital radius R (
, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.
We know that the moon makes a full circumference (
) in 388800 seconds, therefore its tangential velocity is:

where we rounded the velocity to one decimal.
Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.
Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

Answer:
The speed of ejection is 
Solution:
As per the question:
Magnetic field density, B = 0.4 T
Density of the material in the sunspot, 
Now,
To calculate the speed of ejection of the material, v:
The magnetic field energy density is given by:

This energy density equals the kinetic energy supplied by the field.
Thus


where
m = mass of the sunspot in
= 


Answer:
Explanation:
I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.
Felectric = q*E
<span> Ftranslational = m*a
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a
</span><span> Solve for a
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive"
</span><span> q = 1.6*10^-19 C
</span><span> m = 1.67*10^-27 kg
</span><span> E = -150 N/C (- because it is down and up is positive)
</span> a =<span>
-6,4*10^5</span><span> m/s^2 (downward)
</span> answer
a = -6,4*10^5 m/s^2 (downward)