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3 years ago

Basic leads have a probe on one end for making the connection with the electrical circuit being tested and a

Engineering

1 answer:

Ksivusya [100]3 years ago

5 0

Explanation:

mggysluslysltslystksjtskttksnn be my

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In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri

ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

$\sigma_x$ is the normal stress which is 1750 psf
$\sigma_y$ is 0
$\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$

5 0

3 years ago

Select the correct answer.

bonufazy [111]

Answer:

OD.

plate

Explanation:

3 0

3 years ago

If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential

drek231 [11]

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

$\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U$ $15.5=-3.5+0+\bigtriangleup U$

$\bigtriangleup U=19$ kj.

Step3

Specific internal energy is calculated as follows:

$\bigtriangleup u=\frac{\bigtriangleup U}{m}$

$\bigtriangleup u=\frac{19}{5.4}$

$\bigtriangleup u=3.5$ kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

7 0

4 years ago

What character string does the binary ASCII code 1010100 1101000 1101001 1110011 0100000 1101001 1110011 0100000 1000101 1000001

Radda [10]

Answer: This is EASY!

Explanation:

To make it easy, you would convert those binary numbers and to denary. And this gives:

84 104 105 115 32 105 115 32 69 65 83 89 33

Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!

4 0

3 years ago

Energy is not conserved because it can be transferred in term of work and heat. a)-True b)-False

Bess [88]

Answer:

(b) False

Explanation:

Energy is always conserved about energy it is said that energy is never destroyed ,so it is always conserved , energy can only be transferred from one form to another form

For example when we do work the our stored energy is used in doing work means energy is changes from one form to another one more example is that when a ball is falling from any height then its stored potential energy is changes into kinetic energy

4 0

4 years ago