At the initial state: v1 = vf = 0.001053 m
3
/kg, h1 = hf = 467.11 kJ/kg, and s1 = sf = 1.4336 kJ/kgK.
The mass of the water is: m = V/v1 = 0.005/0.001053 = 4.7483 kg.
To find the final state, we will use the First Law:
Q12 = m(h2 - h1) for closed system undergoing a constant pressure process.
h2 = 1Q2/m + h1 = 2200/4.7483 + 467.11 = 930.43 kJ/kg.
At P2 = P1 = 150 kPa, this is a saturated mixture.
hf = 467.11 kJ/kg, hfg = 2226.5 kJ/kg, sf = 1.4336 kJ/kgK, and sfg = 5.7897 kJ/kgK
s2 = sf + sfg (h2 – hf )/hfg = 1.4336 + 5.7897(930.43 – 467.11)/2226.5 = 2.6384 kJ/kgK.
The entropy change of water is:
Delta Ssys= m(s2 – s1) = 4.7483(2.6384 – 1.4336) = 5.72 kJ/K.
Let distance be x.
Time in going, t 1 = x/50
Time in coming, t 2 = x/40
Total time T = x/40 +x/50 = 9x/200
Average speed = total distance/ total time
=( x+x)/(9x/200)= 2x ×200/9x = 400/9 = 44.44 km/h
Answer:
0.304 m/s2
Explanation:
If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.
We can use Newton's 2nd law to calculate the net acceleration of the system
The size of the star defines wither of not it is a dwarf star, or a red giant.
