Answer:
The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Explanation:
Given that,
Initial pressure = 10 bar
Temperature = 22°C
Final pressure = 2.0 bar
We using the value of h
The refrigerant during expansion undergoes a throttling process
Therefore, 
We need to calculate the quality of the refrigerant at the exit of the expansion valve
At 2.0 bar,
The property of ammonia
Using formula
Put the value into the formula
Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Answer:
<h2>
m/s ^2</h2><h2 />
Explanation:
Solution,
When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s
Initial velocity ( u ) = 0 m/s
Final velocity ( v ) = 72 km/h ( Given)
We have to convert 72 km /h in m/s
m/s
Final velocity ( v ) = 20 m/s
Time taken ( t ) = 2 seconds
Acceleration (a) = ?
Now,
we have,
m/s ^2
Hope this helps...
Good luck on your assignment..
Answer:
X: Low potential energy
Y: High Potential energy
Z: Flow of electrons
Explanation: From the figure, it's obvious that Z is the flow of electrons, as shown by the arrow demonstrating the direction of the flow. Because of this, we can easily nullify choices B and C.
From the figure, we can notice that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.
Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is, 
Angle of projection is, 
Time taken to land on the hill is, 
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:
Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.
