Water enters a leaky cylindrical tank (D = 1 ft) at a rate of 8 ft3/min. Water leaks out of the tank at a rate of 17% of the flo
1 answer:
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Answer:
D=41.48 ft
Explanation:
Given that
y=0.5 x²
Vx= 2 t
We know that
At t= 0 ,x=0
At t= 3 s
x= 9 ft
When x= 9 ft then
y= 0.5 x 9² ft
y= 40.5 ft
So distance from origin is
x= 9 ft ,y= 40.5 ft
D=41.48 ft
Vx= 2 t
At t= 3 s , x= 9 ft
y=0.5 x²
y=0.5 x²
Given that
Answer: Hello the question is incomplete below is the missing part
Question: determine the temperature, in °R, at the exit
answer:
T2= 569.62°R
Explanation:
T1 = 540°R
V2 = 600 ft/s
V1 = 60 ft/s
h1 = 129.0613 ( value gotten from Ideal gas property-air table )
<em>first step : calculate the value of h2 using the equation below </em>
assuming no work is done ( potential energy is ignored )
h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778
∴ h2 = 136.17 Btu/Ibm
From Table A-17
we will apply interpolation
attached below is the remaining part of the solution
