1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con
1 answer:
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Answer:
a)Δs = 834 mm
b)V=1122 mm/s
Explanation:
Given that
a)
When t= 2 s
s= 114 mm
At t= 4 s
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
At t= 5 s
V=1122 mm/s
We know that acceleration a
a= 90 t
a = 90 x 5
Answer:
Explanation:
Given
Airline flying at 34,000 ft.
Cabin pressurized to an altitude 8,000 ft.
We know that at standard condition ,density of air
We know that pressure difference
ΔP=ρ g ΔZ
Here ΔZ=34,000-8,000 ft
ΔZ=26,000 ft
ΔP=0.074 x 32.2 x 26,000
So pressure difference will be .