Explain why the following scenario fails to meet the definition of a project description.

Which option identifies the free resource Judi can use in the following scenario?

lakkis [162]

Answer:

Computer programming for three dimensional designs

Explanation:

Doll is not a 2D creation . it's a 3D creation
So on creating the design on 3D scale it's more effective to determine what can be added more.

4 0

2 years ago

Dampness or moisture introduces ____ into the weld, which causes cracking when some metals are welded.

N76 [4]

Answer: Dampness or moisture introduces hydrogen into the weld, which causes cracking when some metals are welded.

Explanation:

<em>This moisture (hydrogen) is a major cause of weld cracking and porosity. </em>

5 0

2 years ago

The air contained in a room loses heat to the surroundings at a rate of 60 kJ/min while work is supplied to the room by computer

charle [14.2K]

Answer:

The net amount of energy change of the air in the room during a 10-min period is 120 KJ.

Explanation:

Given that

Heat loss from room (Q)= 60 KJ/min

Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s

We know that 1 W = 1 J/s

Sign convention for heat and work:

1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.

2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.

So

Q = -60 KJ/min

In 10 min Q = -600 KJ

W = -1.2 KJ/s

We know that

1 min = 60 s

10 min = 600 s

So W = -1.2 x 600 KJ

W = -720 KJ

WE know that ,first law of thermodynamics

Q = W + ΔU

-600 = - 720 + ΔU

ΔU = 120 KJ

The net amount of energy change of the air in the room during a 10-min period is 120 KJ.

4 0

3 years ago

Read 2 more answers

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

blagie [28]

Answer:

0.304 L of Freon is needed

Explanation:

Q = mCT

Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J

C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K

T is temperature in the area of Mars = 189 K

m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg

Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3

Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L

7 0

2 years ago

1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per

KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0

2 years ago