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2 years ago

8

8. Question You should never drive: When weather conditions are not ideal. When road conditions are not ideal. Without having ma

de the necessary
All of the above.

1 answer:

LiRa [457]2 years ago

6 0

The answer is All of the above

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Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air con

PSYCHO15rus [73]

Answer:

See explaination

Explanation:

The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.

Kindly check attachment for the step by step solution of the given problem.

4 0

3 years ago

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?

skad [1K]

Answer:

d

Explanation:

is the because that's the amount of work in making machine can do producing heat

7 0

4 years ago

This is just so I can flip the picture

Pepsi [2]

Answer:

ok

Explanation:

thx for points

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4 years 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

Consider the control volume form of the basic laws. For the conservation of mass form for a control volume with mass flow into a

aleksandrvk [35]

Answer:

C) Dependent on the mass flows in and out.

Explanation:

Lets take control volume(CV)

Take

$m_i$ =inlet mass flow rate

$m_e$ =exit mass flow rate

If we take unsteady flow process then inlet mass can not be equal to exit mass.Some mass can store if inlet mass flow rate is high and exit mass flow rate is low.

So mass of control volume

$m_{cv}=m_i-m_e$ .

so above we can say that mass of control volume dependent on inlet and exit mass.

5 0

3 years ago