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Brilliant_brown [7]
3 years ago
8

Can someone pls give me the answer to this?

Engineering
1 answer:
olganol [36]3 years ago
7 0

I think option c 12 is currect

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The fluid-conditioning components of hydraulic-powered equipment provide fluid that is clean and maintained at an acceptable ope
Korvikt [17]

Answer:Students learn about the fundamental concepts important to fluid power, which includes both pneumatic (gas) and hydraulic (liquid) systems. Both systems contain four basic components: reservoir/receiver, pump/compressor, valve, cylinder.

Explanation:

Students learn about the fundamental concepts important to fluid power, which includes both pneumatic (gas) and hydraulic (liquid) systems. Both systems contain four basic components: reservoir/receiver, pump/compressor, valve, cylinder.

3 0
3 years ago
The removed soil at an excavation site is also called spoil?​
SOVA2 [1]

Answer:

True, That is correct. Soil removed from an excavation site is indeed called spoil.

Spoil definition: The waste material (such as soil) brought up during the course of an excavation.

Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D

4 0
3 years ago
Please help me with this. Picture
qwelly [4]

Answer:

^547.'":8765456\⇆⊇⊇ys p)(ay^547.'":8765456\⇆⊇⊇ys p)(*

iuytrT&*7654567iol;';";lk↓ωω*&65∴∀Hgtre

6и876&*b n™Ο65656&^cxCv876и876&*b n™‰‰ay^547.'":87765^&*  kIKUYtb

Explanation:

=yx^z3

ÜÜÜ

5 0
3 years ago
In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain
swat32

Answer:n=0.973

Explanation:

Given

When True strain\left ( \epsilon _T_1\right )=0.171

at \sigma _1=263.8 MPa

When True stress\left ( \sigma _2\right )=346.2 MPa

true strain \left ( \epsilon _T_2\right )=0.226

We know

\sigma =k\epsilon ^n

where \sigma=True stress

\epsilon=true strain

n=strain hardening exponent

k=constant

Substituting value

263.8=k\left ( 0.171\right )^n------1

346.2=k\left ( 0.226\right )^n-----2

Divide 1 & 2 to get

\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n

1.312=\left ( 1.3216\right )^n

Taking Log both side

ln\left ( 1.312\right )=nln\left ( 1.3216\right )

n=0.973

6 0
3 years ago
A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is
Nataly [62]

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet  = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql  = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

5 0
4 years ago
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