Explanation:
what is the question? could you pls provide it
Answer:
Explanation:
Given that,
The mass of a Hubble Space Telescope, 
It orbits the Earth at an altitude of 
We need to find the potential energy the telescope at this location. The formula for potential energy is given by :
Where
is the mass of Earth
Put all the values,
So, the potential energy of the telescope is 
.
Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.
-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.
-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.
-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.
-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.
-- 'Kilowatt' is a bigger <em>rate</em> of using energy . . . 1,000 joules per second.
-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .
-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .
Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.
Answer:
The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?
Explanation:
3.154x10^7=3.154x10000000=31540000
11.5x31540000=362710000
Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then
Solving for 
we get
Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is
The exhaust rate dm/dt is
Therefore, the velocity at which the exhaust gases exit the engines is
