Answer:
The mass of the massive object at the center of the Milky Way galaxy is 
Explanation:
Given that,
Diameter = 10 light year
Orbital speed = 180 km/s
Suppose determine the mass of the massive object at the center of the Milky Way galaxy.
Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.
We need to calculate the radius of the orbit
Using formula of radius



We need to calculate the mass of the massive object at the center of the Milky Way galaxy
Using formula of mass

Put the value into the formula


Hence, The mass of the massive object at the center of the Milky Way galaxy is 
These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.
Answer:
20.96 h
Explanation:
The perimeter of the track is 2*pi*r = 20pi miles
In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.
From here, we can express the distance of A from X like this:
xa = 3t
And the distance of B would be:
xb = 20pi - 20 - 2t
The time t where they would passed each other and put 12 miles between them would be the one where xa - xb is equal to 12:
xa - xb = 12
3t - (20pi - 20 - 2t) = 12
5t = 20 pi - 8
t = (20pi - 8)/5 = 10.96 h
Remember to add this value to the 10 hours car B had already been racing:
t = 20.96h
Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s
2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.
3) Balanced. Both boys are pulling with the same force. Neither is winning.
4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.
5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.
6) Balanced. You and the dog are pulling with the same force.