Answer:
Geothermal energy can heat, cool, and generate electricity: Geothermal energy can be used in different ways depending on the resource and technology chosen—heating and cooling buildings through geothermal heat pumps, generating electricity through geothermal power plants, and heating structures through direct-use
Explanation:
Answer:
Explanation:
Calculating the exit temperature for K = 1.4
The value of
is determined via the expression:

where ;
R = universal gas constant = 
k = constant = 1.4


The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :
------ equation(1)
we can rewrite the above equation as :


where:



Thus, the exit temperature = 402.36 K
The exit pressure is determined by using the relation:



Therefore, the exit pressure is 17.79 bar
Acceleration is the rate at which an object changes its velocity. It defines how much the velocity is changing. The acceleration can be negative and positive. Negative acceleration is when the object slows down, while positive while the object goes faster.
<span>A ball moving at a constant speed around a circular track produces acceleration. </span>
Answer:
5 years worth of work (aka all of the homework i currently have)