1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is

Therefore, the angular velocity of the telescope is

2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is

Therefore, the linear velocity of the telescope is:

Answer:When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor.
Explanation:This is what I would say is the answer bc I had to do reasearch on a lot of this for my work this year so if its not im veery sorry
Water breaks down into hydrogen and oxygen: It is a decomposition reaction as a single substance decomposes to give two products.

Leaves make starch using chlorophyll and carbon dioxide: Synthesis reaction: as the synthesis reaction involves two or more than two reactants which join together to result into a single main product along with the formation of simple by products.

Food burns in oxygen gas and releases a lot of energy: Combustion: Combustion process involves the use of oxygen to give products along with release of energy.

Adding vinegar (acid) to baking soda (alkali) gives a product that is neither acidic nor alkaline: Neutralization: acetic acid in vinegar reacts with soda (base) to give salt (neutral) and .

Answer:
The net gravitational force on the mass is 
Explanation:
We have by Newton's law of gravity the force of attraction between masses 

Applying vales we get
Force of attraction between 135 kg mass and 38 kg mass is

Force of attraction between 435 kg mass and 38 kg mass is

Thus the net force on mass 38.0 kg is 