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: freemasonry is Being a Mason is about a father helping his son make better decisions; a business leader striving to bring morality to the workplace; a thoughtful man learning to work through tough issues in his life.
Explanation:
Answer:![121\ \Omega](https://tex.z-dn.net/?f=121%5C%20%5COmega%20)
![0.909\ A](https://tex.z-dn.net/?f=0.909%5C%20A)
Explanation:
Given
Power ![P=100\ W](https://tex.z-dn.net/?f=P%3D100%5C%20W)
Voltage applied ![V=110\ V](https://tex.z-dn.net/?f=V%3D110%5C%20V)
Resistance of the bulb is given by
![P=\frac{V^2}{R}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BV%5E2%7D%7BR%7D)
![100=\frac{110^2}{R}](https://tex.z-dn.net/?f=100%3D%5Cfrac%7B110%5E2%7D%7BR%7D)
![R=\frac{12100}{100}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B12100%7D%7B100%7D)
![R=121\ \Omega](https://tex.z-dn.net/?f=R%3D121%5C%20%5COmega%20)
Current drawn by the Power source is given by
![P=V\cdot I](https://tex.z-dn.net/?f=P%3DV%5Ccdot%20I)
![I=\frac{P}{V}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BP%7D%7BV%7D)
![I=\frac{100}{110}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B100%7D%7B110%7D)
![I=0.909\ A](https://tex.z-dn.net/?f=I%3D0.909%5C%20A)
Answer:
Lay the person down and elevate thier legs slightly.
Explanation:
Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.
Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:
![P_{total} = P_{1}+P_{2}+...](https://tex.z-dn.net/?f=P_%7Btotal%7D%20%3D%20P_%7B1%7D%2BP_%7B2%7D%2B...)
The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:
= moles of a gas / total number moles of gas
The rigid tank has total pressure of 1MPa.
molar mass = 14g/mol
mass in the tank = 2000g
number of moles in the tank:
= 142.85mols
molar mass = 44g/mol
mass in the tank = 4000g
number of moles in the tank:
= 90.91mols
Total number of moles: 142.85 + 90.91 = 233.76 mols
To calculate partial pressure:
![P_{i}=P_{total}.X_{i}](https://tex.z-dn.net/?f=P_%7Bi%7D%3DP_%7Btotal%7D.X_%7Bi%7D)
For Nitrogen gas:
![P_{N_{2}}=1.\frac{142.85}{233.76}](https://tex.z-dn.net/?f=P_%7BN_%7B2%7D%7D%3D1.%5Cfrac%7B142.85%7D%7B233.76%7D)
= 0.6
For Carbon Dioxide:
![P_{total}=P_{N_{2}}+P_{CO_{2}}](https://tex.z-dn.net/?f=P_%7Btotal%7D%3DP_%7BN_%7B2%7D%7D%2BP_%7BCO_%7B2%7D%7D)
![P_{CO_{2}} = P_{total}-P_{N_{2}}](https://tex.z-dn.net/?f=P_%7BCO_%7B2%7D%7D%20%3D%20P_%7Btotal%7D-P_%7BN_%7B2%7D%7D)
![P_{CO_{2}}=1-0.6](https://tex.z-dn.net/?f=P_%7BCO_%7B2%7D%7D%3D1-0.6)
0.4
Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.