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3 years ago

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A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surfac

e is 5 m. The atmospheric pressure is 95 kPa. Considering the atmospheric pressure, the hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.

1 answer:

UkoKoshka [18]3 years ago

3 0

Answer:

864 KN

Explanation:

Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.

Please kindly check attachment for the step by step solution of the given problem.

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How does running an electric current through wire cause a magnetic field?

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Answer:

When a charged particle—such as an electron, proton or ion—is in motion, magnetic lines of force rotate around the particle. Since electrical current moving through a wire consists of electrons in motion, there is a magnetic field around the wire.

Explanation:

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I really need help i will give brainly plz no funny answers

vfiekz [6]

Answer:

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3 years ago

Nitrogen gas is compressed at steady state from a pressure of 14.2 psi and a temperature 60o F to a pressure of 120 psi and a te

brilliants [131]

Answer:

a) 229.4281 hp.

b) 262.15 ft3/min.

Explanation:

Given data:

P1 = 14.2 psi

T1 = 60°F = 520° R

P2 = 120 psi

T2 = 500°F = 960° R

volumetric flow rate ( Av1 ) = 1200 ft^3 /min = 20 ft^3 / sec

attached below is the detailed solution

5 0

3 years ago

How many grams of perchloric acid, HClO4, are contained in 37.6 g of 70.5 wt% aqueous perchloric acid? How many grams of water a

kap26 [50]

Answer :

The mass of perchloric acid is, 26.5 grams.

The mass of water in the same solution is, 11.1 grams

Explanation :

As we are given that 70.5 wt % aqueous perchloric acid that means 70.5 grams of perchloric acid present in 100 grams of solution.

Now we have to determine the mass of perchloric acid in 37.6 grams of aqueous perchloric acid.

As, 100 grams of aqueous perchloric acid (solution) contains 70.5 grams of perchloric acid.

So, 37.6 grams of aqueous perchloric acid (solution) contains $\frac{37.6}{100}\times 70.5=26.5$ grams of perchloric acid.

Thus, the mass of perchloric acid is, 26.5 grams.

Now we have to determine the mass of water are in the same solution.

Total mass of solution = 37.6 g

Mass of perchloric acid = 26.5 g

Mass of water = Total mass of solution - Mass of perchloric acid

Mass of water = 37.6 g - 26.5 g

Mass of water = 11.1 g

Thus, the mass of water in the same solution is, 11.1 grams

4 0

3 years ago

Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase

ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ( $\% V_{Gr}$ ) is determined by the following expression:

$\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%$

Where:

$V_{Gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{Fe}$ - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

$V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}$

$V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}$

Where:

$m_{Gr}$ , $m_{Fe}$ - Masses of the graphite and ferrite phases, measured in grams.

$\rho_{Gr}$ , $\rho_{Fe}$ - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

$\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%$

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

$m_{Gr} = \frac{2.5}{100}\times (100\,g)$

$m_{Gr} = 2.5\,g$

$m_{Fe} = 100\,g - 2.5\,g$

$m_{Fe} = 97.5\,g$

If $m_{Gr} = 2.5\,g$ , $m_{Fe} = 97.5\,g$ , $\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}$ and $\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}$ , the volume percent of graphite is:

$\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%$

$\% V_{Gr} = 91.906\,\%$

The volume percent of graphite is 91.906 per cent.

6 0

3 years ago