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1 answer:
Answer:
Explanation:
Initially the packet was ascending up with the balloon.
Taking upward as positive direction;
initial velocity, u = 4.9 m/s
final velocity = v m/s
initial height, h₁ = 100 m
final height, h₂ = 0
a = -9.8 m/s²
time taken = t seconds
Considering the second equation of motion.
S = ut + 1/2at^2
s = ut + 0.5at²
(h₂-h₁) = ut + 0.5at²
0- 100 = 4.9t + 0.5×(-9.8)×t²
-100 = 4.9t - 4.9t²
4.9t² -4.9t - 100 =0
t² -t - 20.41 = 0
Using : b±√(b²-4ac))/(2a).... 1
Where a =1, b = -1 c =-20.408
Substituting the values into equation 1
Solving it, we get t = 5.045s
v = u + at = 4.9 -9.8×5.045 = 4.9 - 49.44 = -44.54 m/s
The final velocity is -44.54 m/s, and it took 5.045s to reach the ground.
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Answer:
Hi
Final temperature = 250.11 °C
Final volume = 0,1 m3.
Process work = 0
Explanation:
The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.
This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:
v=0.05 m3/kg
v1=0.057061 m3/kg
T1=242.56°C
v2=0.049779 m3/kg
T2=250.35°C
T=
The process work is zero because there is no change in volume during heating:
W=PxΔv=Px0=0
where
W=process work
P=pressure
Δv=change of volume, is zero because the piston was blocked so the volume remains constant.