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vekshin1
3 years ago
7

You guys are amazing :D

Engineering
2 answers:
iren [92.7K]3 years ago
4 0
Thank you. You’re amazing as well !
Sloan [31]3 years ago
3 0
Ik i am thank you tho xoxo
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An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec
Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0
3 years ago
A rectangular workpiece has the following original dimensions: 2a=100mm, h=25mm, and width=20mm. The metal has a strengh coeffic
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Answer:

See attachment for detailed answer.

Explanation:

Download pdf
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3 years ago
Why the inviscid, incompressible, and irrotational fields are governed by Laplace's equation?
creativ13 [48]

Answer: Laplace equation provides a linear solution and helps in obtaining other solutions by being added to various solution of a particular equation as well.

Inviscid , incompressible and irrotational field have and basic solution ans so they can be governed by the Laplace equation to obtain a interesting and non-common solution .The analysis of such solution in a flow of Laplace equation is termed as potential flow.

6 0
3 years ago
A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o
navik [9.2K]

The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

useful output energy = 2400 J

Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

h = 2400/800

h = 3 meters

Read more about Machine Efficiency at; brainly.com/question/3617034

#SPJ1

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2 years ago
I have a 308 bullet stuck in my ak,s mag help?
Assoli18 [71]
Lick the bullet and push it down
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3 years ago
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