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Verizon [17]
4 years ago
12

Check the answer that best describes the relationship between f(x) and x. (For example if f(x) is Θ(x) check that as your answer

and not O(x) or Ω(x) even though these are true also in this case.)
(a) f(x)=10 is

A. O(x) O(x)B. Ω(x) Ω(x)C. Θ(x) Θ(x)
Engineering
1 answer:
egoroff_w [7]4 years ago
3 0

Answer:

Correct option is A - O(x)

Explanation:

Given f(x) = 10 and g(x) = x

Hence f(x) is O g(x)

As such correct answer is O(x) which is option A

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Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph
kicyunya [14]

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:

S.G=\frac{\rho_{sw}}{\rho_{w}}

Here, density of sea water is\rho_{sw}, surface gravity is S.G and density of water is \rho_{w}.

Substitute all the values in the above equation as follows:

S.G=\frac{\rho_{sw}}{\rho_{w}}

1.025=\frac{\rho_{sw}}{1000}

\rho_{sw}=1025 kg/m³.

Step2

Difference in pressure is calculated as follows:

\bigtriangleup p=rho_{sw}gh

\bigtriangleup p=1025\times9.81\times320

\bigtriangleup p=3217680 pa.

Or

\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})

\bigtriangleup p=3217.68 kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0
3 years ago
A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu
Ksju [112]

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

7 0
3 years ago
malott, m. e. (2003). paradox of organizational change: engineering organizations with behavioral systems analysis.
mylen [45]

Answer:

Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.

7 0
2 years ago
As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de
katen-ka-za [31]

yes it will

Explanation:

5 0
3 years ago
All of these are true about GMA (MIG) welding EXCEPT that:
Hatshy [7]

Answer:

the welding gun liner regulates the shielding gas.

Explanation:

The purpose of the welding gun liner is to properly position the welding wire from the wire feeder till it gets to the nozzle or contact tip of the gun. <em>Regulation of the shielding gas depends on factors such as the speed, current, and type of gas being used. </em>In gas metal arc welding, an electric arc is used to generate heat which melts both the electrode and the workpiece or base metal.

The electric arc produced is shielded from contamination by the shielding gas. The heat generated by the short electric arc is low.

3 0
3 years ago
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