Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg
Answer:
85 N backwards
Explanation:
u subtract then thats the net force and add direction
Explanation:
It is given that,
Momentum of an object, p = 23.3 kg-m/s
Kinetic energy, E = 262 J
(a) Momentum is given by, p = mv
23.3 = mv...........(1)
Kinetic energy is given by, 
m = mass of the object
v = speed of the object
v = 22.48 m/s
(2) Momentum, p = mv
m = 1.03 Kg
Hence, this is the required solution.
1 to 2 seconds because it had the steepest slope
