Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:

The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
So,
GPE (graviational potential energy) = mass x g x height
GPE is depends on where zero height is defined. In this situation, we define h = 0 as the initial height.



The builder has gained 18.375 kJ of PE.
Answer:
A- mass and type of material
B- type of material
C- Temperature
Explanation:
thx
To develop this problem it is necessary to apply the concepts related to the Dopler effect.
The equation is defined by

Where
= Approaching velocities
= Receding velocities
c = Speed of sound
v = Emitter speed
And

Therefore using the values given we can find the velocity through,


Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1


Therefore the cars goes to 145.3m/s
Answer:
Tension T1 is less than tension T2.
T1 < T2
Explanation:
According to given data,
mass of box A ( mA) is grater than mass of box B (mB)
we can write,
m(A) > m(B)
Newton's second law states that:
Tension of object is directly proportional to the mass of the system.
T ∝ m
here Boxes A and B are being pulled to the right on a frictionless surface,
so Tension T1 generates due to the mass of box A m(A)
and Tension T2 arises due to mass of the system m(A) + m(B)
Thus tension T1 will be less than tension T2
T1 < T2
learn more about Tension force here:
<u>brainly.com/question/13175014</u>
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