Answer: I think it's C
Explanation: I hope this helps (Sorry if it doesn't)
Answer:
oceans is the answer that I got
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
In response of what like what’s the full clear question
Answer:
166,600J
Explanation:
Kinetic energy (K.E), which is the energy due to motion of a body, can be calculated by using the formula;
K.E = 1/2 × m × v²
Where;
K.E = kinetic energy (joules)
m = mass of body (kg)
v = speed or velocity (m/s)
According to this question, the mass of the roller coaster is 833.0 kg while its velocity/speed is 20.0m/s.
K.E = 1/2 × 833 × 20²
K.E = 1/2 × 833 × 400
K.E = 1/2 × 333200
K.E = 166,600
Therefore, the kinetic energy of the roller coaster car is 166,600J.
