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2 years ago

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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masya89 [10]

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2 years ago

Inspections may be_____ or limited to a specific area such as electrical or plumbing

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3 years ago

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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