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Rom4ik [11]
3 years ago
9

What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Physics
1 answer:
12345 [234]3 years ago
5 0

Incomplete question.The complete question is

A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Answer:

Total gauge pressure= 3724Pa or 3.724kPa

Explanation:

Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.

Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.

Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa

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mrs_skeptik [129]

Answer:

Current through each phase Vp = 2.2A

Total three phase power Pt= 1.45kW

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Explanation:

i) Find current through each phase

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Z =100 Ω

I = Vp/Z

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ii) Find the total three phase power

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Power for each phase is given as:

P = 220 * 2.2

  = 484 W

Total power TP =3* P

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 iii) Find the power factor of the load

Phase angle for a resistive load is 0.

α= 0

Hence, power factor of load = cos α

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5 0
3 years ago
Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared
Aloiza [94]

Answer:

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Explanation:

Given:

The area of the house A = 2050\ ft^{2}

The height of the house h=10\ ft

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

Volume = Area\times height

Volume = A\times h

Area and height of the house are known, so we substitute these value in above equation.

Volume = 2050\times 10

Volume = 20500\ ft^{3}

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for m^{3}

Volume = \frac{20500}{3.2808}

Volume = 6248.48\ m^{3}

Therefore, the volume of the house is 6248.48 m^{3}

8 0
3 years ago
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Ksenya-84 [330]

Answer:

It looks like its moving north.

Explanation:

3 0
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anzhelika [568]

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