Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
![v_1 = vf + x(vg - vf)](https://tex.z-dn.net/?f=v_1%20%3D%20vf%20%2B%20x%28vg%20-%20vf%29)
![= 0.001053 + 0.25(1.1594 - 0.001053)](https://tex.z-dn.net/?f=%3D%200.001053%20%2B%200.25%281.1594%20-%200.001053%29%20)
![v_1 = 0.20964 m^3/kg](https://tex.z-dn.net/?f=v_1%20%3D%200.20964%20%20m%5E3%2Fkg)
![s_1 = Sf + x(Sfg)](https://tex.z-dn.net/?f=s_1%20%3D%20Sf%20%2B%20x%28Sfg%29)
![= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K](https://tex.z-dn.net/?f=%3D%201.4337%20%2B%200.25%20%5Ctimes%205.7894%20%3D%202.88%20kJ%2Fkg%20K)
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so![v_1 - v_2 = 0.29064 m^3/kg](https://tex.z-dn.net/?f=%20v_1%20-%20%20v_2%20%20%3D%200.29064%20m%5E3%2Fkg)
![v_2 = v_1 = vf + x(vg - vf)](https://tex.z-dn.net/?f=v_2%20%3D%20v_1%20%3D%20vf%20%2B%20x%28vg%20-%20vf%29)
![=0.29064 = x_2(0.88578 - 0.001061)](https://tex.z-dn.net/?f=%3D0.29064%20%3D%20x_2%280.88578%20-%200.001061%29)
![x_2 = 0.327](https://tex.z-dn.net/?f=x_2%20%3D%200.327)
![s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K](https://tex.z-dn.net/?f=s_2%20%3D%201.5302%20%2B%200.32%20%5Ctimes%205.5968%20%3D%203.36035%20kJ%2Fkg%20K)
Change in entropy ![\Delta s = m(s_2 - s_1)](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20m%28s_2%20-%20s_1%29)
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice? is mathematically given as
x = 9.93 g
<h3>What is the amount of water that freezes onto the ice?</h3>
Where
Energy received = energy given out
Generally, the amount of water is mathematically given as
(53)(0.5)(30) = (80)(x)
Therefore
x = (49)(0.5)(16)/(80)
x = 9.93 g
In conclusion, the mass of water
x = 9.93 g
Read more about mass
brainly.com/question/15959704
Answer:
D. 24 lb
Explanation:
Tina has been dieting for 13 weeks
First week she lost 3 pounds
Next week she gained 1 pound and did not lose any. This will be subtracted as she has gained a pound
The remaining 11 weeks she lost 2 pounds per week
Weight lost in the 11 weeks = 11×2 =22 pounds
Total weight lost
3-1+22 = 24 lb
Tina has lost 24 pounds in total during the 13 weeks
The answer yr looking for would b true!