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2 years ago

4. Which 2D shape on the left would be used to make the 3D shape on the right? (1 pt.)

Engineering

1 answer:

dexar [7]2 years ago

7 0

it would be a bc its a sqare?well 3d is like you can say a cube 2d is like flat

Explanation:

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primitive transportation and storage systems that make local distribution ineffective if not impossible, the lack of clean water

Vera_Pavlovna [14]

Primitive transportation and storage systems that make local distribution ineffective if not impossible, and the lack of clean water and the lack of effective sewer systems are all examples of the barrier called physical & environmental barriers.

<h3>What are the problems with having physical & environmental barriers?</h3>

Time, Place, Space, Climate, and Noise are the key environmental/physical constraints. Some are simple to change, while others may prove to be difficult.

Physical barriers are essentially defined by three primary factors: environment, distance, and medium unawareness. Environmental barriers, on the other hand, are connected to variables that occur in the current environment.

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1 year ago

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

If you are involved in a collision where there is injury, you must report the incident within .......

Misha Larkins [42]

Answer:

24 hours

Explanation:

you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.

4 0

3 years ago

Tech A says that serviceable wheel bearings can be repacked by removing the dust cap, filling it with grease, and reinstalling i

rewona [7]

Serviceable wheel bearings can be repacked by removing the dust cap and the cotter pin must be replaced with a new one every time it is removed. Therefore, both Tech A and B are correct.

Sealed bearing assemblies are typically prefilled with lubricant. They also make a more reliable and consistent installation process due to the fact that every into comes preset at the proper clearance.

Furthermore, the serviceable wheel bearings can be repacked by removing the dust cap, filling it with grease, and reinstalling it. Also, cotter pin must be replaced with a new one every time it is removed.

Therefore, both tech A and B are correct.

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2 years ago

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t

iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, $P_{1}$ = 1 bar

$v_{2}$ = 4 bar

pV¹°¹ = constant

From superheated propane table, at $P_{1}$ = 1 bar and $T_{1}$ = 27⁰C

$v_{1}$ = 0.557 m³/kg

$v_{2}$ = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

$V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1 \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557 \\\\V2 = 0.158 m^3/kg$

At = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

$W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\ \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\ \\W = -19.75KJ$

(b) Heat transfer = ?

$Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ$

8 0

4 years ago