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Mashutka [201]
2 years ago
6

How are sedimentary - gravity flows different from fluid - gravity flows

Engineering
1 answer:
Nimfa-mama [501]2 years ago
3 0

Answer:

Sediment gravity flows are mixtures of water and sediment particles where the gravity acting on the sediment particles moves the fluid, in contrast to rivers, where the fluid moves the particles.

Explanation:

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Imagine you are a process safety consultant and you have been tasked to make a metal refinery site DSEAR compliant. What are the
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Complying with DSEAR involves:

Assessing risks. ...

Preventing or controlling risks. ...

Control measures. ...

Mitigation. ...

Preparing emergency plans and procedures. ...

Providing information, instruction and training for employees. ...

Places where explosive atmospheres may occur ('ATEX' requirements)

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Inspections may be_____ or limited to a specific area such as electrical or plumbing
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A is the answer for the sentence
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What is made in heaven?​
kramer

Answer:

Babies come from heaven didn't you know?

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2 years ago
A rotating beam is subject to an alternating stress of 48 kpsi and a mean stress of 24 kpsi. The ultimate strength of the materi
Alisiya [41]

Answer:

goodman = 0.694

life of beam = 211597

Explanation:

alternating stress = 48 kpsi

mean stress = 24 kpsi

ultimate strength = 100 kpsi

endurance limit = 40 kpsi

goodman:

= \frac{mean stress}{ultimate stress} +\frac{alternating stress}{endurance limit} =\frac{1}{N}

= \frac{24}{100} +\frac{48}{40} =\frac{1}{N}

= 0.24 + 1.2 = \frac{1}{N}

N = 1/1.44

N = 0.694

2. check attachment for diagram

Log(N)-3/3 = log90 - log48/log90 - log40

Log(N)-3/3 = 0.77517

Log N = 5.325509

N = 10^(5.325509)

N = 211597

6 0
2 years ago
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin
Leni [432]

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

6 0
2 years ago
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