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:
1.8 m/s
Explanation:
Draw a free body diagram of the block. There are four forces:
Normal force Fn up.
Weight force mg down.
Applied force F to the east.
Friction force Fn μ to the west.
Sum the forces in the y direction:
∑F = ma
Fn − mg = 0
Fn = mg
Sum the forces in the x direction:
F − Fn μ = ma
F − mg μ = ma
a = (F − mg μ) / m
a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg
a = 0.53 m/s²
Given:
Δx = 3 m
v₀ = 0 m/s
a = 0.53 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)
v = 1.8 m/s
Answer: 0.5 m
Explanation:
Given
Mass of the person is 
Trampoline launches the person into the air up to height of 
Force experience by springs is 
Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.
![\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m](https://tex.z-dn.net/?f=%5CRightarrow%20F%5Ccdot%20x%3Dmgh%5Cquad%20%5B%5Ctext%7Bx%3Ddisplacement%20of%20the%20trampoline%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B50%5Ctimes%209.8%5Ctimes%202%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B980%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D0.5%5C%20m)
Answer:
The periodic table illustrate some of the elements from Hydrogen to Calcium
