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2 years ago

A tank contains 500 kg of a liquid whose specific gravity is 2. Determine the volume of the liquid in the tank.

Engineering

1 answer:

horsena [70]2 years ago

4 0

Answer:

$So\ the\ volume\ of\ liquid\ = 0.25\ m^3$

Explanation:

given that

Mass m = 500 kg

Specific gravity = 2

$So\ density\ of\ liquid\ =2000 \frac{kg}{m^3}$

We know that

mass = density x volume

So by putting the values

mass = density x volume

500 = 2000 x volume

$So\ the\ volume\ of\ liquid\ = 0.25\ m^3$

You might be interested in

In Lab 7, we worked through a program that displayed the homeless shelter occupancy over time. The same approach can be used for

Bezzdna [24]

Answer:

Explanation:

The python code to generate this is quite simple to run.

i hope you understand everything written here, you can as well try out other problems to understand better.

First to begin, we import the package;

Code:

import pandas as pd

import matplotlib.pyplot as plt

name = input('Enter name of the file: ')

op = input('Enter name of output file: ')

df = pd.read_csv(name)

df['Date'] = pd.to_datetime(df["Date"].apply(str))

plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")

plt.legend(loc="upper right")

plt.xticks(rotation=20)

plt.savefig(op)

plt.show()

This should generate the data(plot) as seen in the uploaded screenshot.

thanks i hope this helps!!!

6 0

2 years ago

Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe

lys-0071 [83]

Answer:

Dalton

The part of the close-out report that would describe how he would plan and manage projects in the future is:

summary of project management effectiveness

Explanation:

The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.

4 0

2 years ago

An aircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa1m (36.4 ksi1in.)

sergiy2304 [10]

Answer:

$Y = \frac{40 Mpa \sqrt{m}}{300 Mpa \sqrt{\pi 0.002m}}= 1.682$

Now we can solve for the fracture level with this formula:

$\sigma_F = \frac{K_{lc}}{Y \sqrt{\pi d_x}}$

We can use the parameter of Y from the previous equation since this value not changes.

$d_x = 6/2 = 3 mm =0.003m$

And replacing we have:

$\sigma_F= \frac{40 Mpa \sqrt{m}}{1.682 \sqrt{\pi 0.003m}}= 244.96 Mpa$

And for this case we see that this value not correspond to 260 Mpa the value reported, so then the correct answer for this case would be 244.96 Mpa for the new crack length of 6mm = 0.006 m

Explanation:

We need to solve for the parameter Y given by this expression:

$Y = \frac{K_{lc}}{\sigma \sqrt{\pi d}}$

Where:

$K_{lc} = 40 MPa \sqrt{m}$ represent the fracture toughness

$\sigma = 300 MPa$ represent the stress level

$d = \frac{4 mm}{2}= 2mm = 0.002 m$

And if we replace we got:

$Y = \frac{40 Mpa \sqrt{m}}{300 Mpa \sqrt{\pi 0.002m}}= 1.682$

Now we can solve for the fracture level with this formula:

$\sigma_F = \frac{K_{lc}}{Y \sqrt{\pi d_x}}$

We can use the parameter of Y from the previous equation since this value not changes.

$d_x = 6/2 = 3 mm =0.003m$

And replacing we have:

$\sigma_F= \frac{40 Mpa \sqrt{m}}{1.682 \sqrt{\pi 0.003m}}= 244.96 Mpa$

And for this case we see that this value not correspond to 260 Mpa the value reported, so then the correct answer for this case would be 244.96 Mpa for the new crack length of 6mm = 0.006 m

6 0

2 years ago

The molecular weight of a 10g rubber band

Archy [21]

Answer:

The answer to this question is 1 * 10g /mole = 10

Explanation:

From the question given, the first step to take is to find the weight of a 10g rubber band.

Now,

The molecular weight of rubber band , tires the primary chains may have molecular weights is of order of 1 x 10 g/

Therefore it is 1 * 10g/mole =10

8 0

2 years ago

Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No

Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is $1.557\times10^{-7}$ m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

$Re=\frac{vd}{\nu}$

Here, v is velocity, $\nu$ is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

$Re=\frac{vd}{\nu}$

$Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}$

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

5 0

2 years ago