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MakcuM [25]
3 years ago
9

Which option identifies the step skipped in the following scenario?

Engineering
2 answers:
Irina-Kira [14]3 years ago
5 0

Answer:

He did not pay attention to the cost analysis of the project.

Explanation:

- Massmaster34

olga nikolaevna [1]3 years ago
4 0

Answer:

he didn't use the proper material

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A rotator has a weight of 100lb with a centroidal radius of gyration of 9 in. Determine the moment of inertia about the center o
Ronch [10]

Answer:

8100 lbin²

Explanation:

Moment or inertia is expressed using the formula

I = mr²

M is the mass of the body

r is the radius of gyration

Given

W = 100lb

r = 9in

Required

Moment of inertia

I = Wr²

I = 100(9)²

I = 100×81

I = 8100lbin²

Hence the moment of inertia about the center of gravity for the rotator is

8100 lbin²

8 0
3 years ago
A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.
Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

E_{in}- E_{out}=E_{system,final}- E_{system,initial}

Where

E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

E_{in}- E_{out}=-Q_{out}+W_{in}

Regarding the energy system we have the following:

E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}

By doing the calculations we have:

E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0
3 years ago
Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i
DedPeter [7]

Answer:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

Explanation:

The internal energy is equal to:

dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV

The entropy is equal to:

dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }

Replacing:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

4 0
3 years ago
A 2-m^3 rigid tank contains nitrogen gas at 500kPa and 300K. Now the gas is compressed isothermally to a volume of 0.1 m. The wo
stira [4]

Answer:

(d) 2996 kJ

Explanation:

We have given initial volume v_1=2m^3

initial pressure p_1=500\ kPa

initial temperature T_1=300\ K

we know that during  isothermal process the work done is given by

W=p_1v_1ln\frac{v_2}{v_1}

where v_2= final\ volume =0.1m^3\ given

putting all these value in formula of work done

W=500\times 2\times ln\frac{0.1}{2}

=-2995.8 kJ here negative sign indicates that work is dine on the gas

so wok done =2995.8 kJ

8 0
3 years ago
Marco is an Italian architect. He has recieved a contract
kirill [66]

Answer:

Marco is an Italian architect. He has received a contract to design a spacious building.

Explanation:

ect.

8 0
3 years ago
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