Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Answer:
11.541 mol/min
Explanation:
temperature = 35°C
Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa
note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )
from steam table it is = 5.6291 Kpa
calculate the mole fraction of H
( YH )
= 5.6291 / 151.95
= 0.03704
calculate the mole fraction of air ( Yair )
= 1 - mole fraction of water
= 1 - 0.03704 = 0.9629
Now to determine the molar flow rate of water vapor in the stream
lets assume N = Total molar flow rate
NH = molar flow rate of water
Nair = molar flow rate of air = 300 moles /min
note : Yair * n = Nair
therefore n = 300 / 0.9629 = 311.541 moles /min
Molar flowrate of water
= n - Nair
= 311.541 - 300 = 11.541 mol/min
Answer:
The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
Explanation:
The distance that the truck starts slowing down = 80 ft from the stop sign
Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.
u = initial velocity of the truck = 40 mph = 58.667 ft/s
v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)
x = horizontal distance covered during the deceleration
a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration
v² = u² + 2ax
0² = 58.667² + 2(-12)(x)
24x = 3441.816889
x = 143.41 ft
143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
Answer:
option (a)
Explanation:
t = 5 sec, α = 2 rad/s^2, f0 = 20 rpm = 20 / 60 rps
Use second equation of motion for rotational motion
θ = ω0 x t + 1/2 α t^2
θ = 2 x 3.14 x 5 x 20 / 60 + 0.5 x 2 x 5 x 5
θ = 10.47 + 25 = 35.47 rad
Number of revolution = 35.47 / (2 x 3.14) = 5.65
