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4 years ago

Please help meeeee. I don’t have time

Engineering

1 answer:

Sholpan [36]4 years ago

3 0

Answer: precision

Explanation: because accuracy is right on there and precision is getting closer and closer

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100000000000x1000000000000=?

zhannawk [14.2K]

Answer:

Thy answer to your very sophisticated question is 1E23

Explanation:

IT JUST IS! Dont ask any questions

mAsquErade, mAsquerade tHat iS mY naME

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3 years ago

Read 2 more answers

Gn bbys sleep well gn

Musya8 [376]

Answer:

mwuah

Explanation:

I got u girl, what r u on?

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3 years ago

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The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate

Sonja [21]

Answer:

a) Mechanical efficiency ( $\varepsilon$ )=63.15% b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, $Wmec=0.95*6kW=5.7 kW$ .

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump $\Delta P$ . So $P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W$ .

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is $\varepsilon=3.6kW/5.7kW=0.6315=63.15\%$

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling H to the heat transfered to the fluid, the specif heat of the water and $\rho$ the density of the water:

$[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T$ [/tex]

Finally, the temperature rise is:

$\Delta T=2100/75348 \ºC=0.028 \ºC$

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3 years ago

On some engines, after torquing cylinder head fasteners, you

djverab [1.8K]

Answer:

Re-torque the bolts as required while your engine is warm. But if you're using aluminum cylinder heads, you should wait until your engine is complete cooled until re-torquing

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3 years ago

Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co

butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

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3 years ago