Answer:
The pressure difference across hatch of the submarine is 3217.68 kpa.
Explanation:
Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.
Given:
Height of the hatch is 320 m
Surface gravity of the sea water is 1.025.
Density of water 1000 kg/m³.
Calculation:
Step1
Density of sea water is calculated as follows:
Here, density of sea water is
, surface gravity is S.G and density of water is 
.
Substitute all the values in the above equation as follows:
kg/m³.
Step2
Difference in pressure is calculated as follows:
pa.
Or
kpa.
Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.
Answer:
Context
Explanation:
It is of great value for an engineer to keep the context of his/her experiment in mind.
Explanation:
a) The total volume equals the sum of the volumes.
500 = x + y
The total octane amount equals the sum of the octane amounts.
89(500) = 87x + 92y
44500 = 87x + 92y
b) desmos.com/calculator/ekegkzllqx
As x increases, y decreases.
c) Use substitution or elimination to solve the system of equations.
44500 = 87x + 92(500−x)
44500 = 87x + 46000 − 92x
5x = 1500
x = 300
y = 200
The required volumes are 300 gallons of 87 gasoline and 200 gallons of 92 gasoline.
Answer:
The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.
