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:
Companies are combining their online business activities with their existing physical presence in order to lower costs of their operations. When both these things are combined labor costs are reduced because with online presence the company has to have limited number of branches, inventory costs are reduced because additional inventories for every physical outlet is not required and delivery costs are reduced because now company don't have to supply the things to all the outlets on regular basis.
Trust of the people is also improved because mostly people are reluctant to order from the brands that only have their online store and donot have any physical presence. Value added services are provided by a company who have both online and offline presence like home delivery and customized offerings.
Answer:
a) The final equilibrium temperature is 83.23°F
b) The entropy production within the system is 1.9 Btu/°R
Explanation:
See attached workings
Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:

where;
fracture toughness K = 137 MPa
geometry factor Y = 1
applied stress
= ???
crack length a = 2mm = 0.002
∴




Now, the tensile impact obtained is:

P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN