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Rainbow [258]
3 years ago
6

Which option demonstrates when most vehicles lose their efficiency?

Engineering
2 answers:
faltersainse [42]3 years ago
7 0

Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

guapka [62]3 years ago
6 0

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

If u think this helped u ..plz mark me as brainliest ..

And follow me

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I need help with my autos
oksano4ka [1.4K]

Answer:

what is wrong with it and what is the question

Explanation:

6 0
3 years ago
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
3 years ago
Your uncle has given you a newmonitor for your computer. When you attempt to connect it, you notice that none of the ports on th
juin [17]

Answer:

Go on amazon and look for HDMI -> (VGA/DVI/HDMI) what ever you need adapters.

6 0
3 years ago
The activation energy for diffusion in BCC iron is 84 kJ/mol. Which of the following explains why the carbide precipitation acti
k0ka [10]

Answer:

  • B. Precipitation require the diffusional activation energy plus an additional energy to form the precipitate.

Explanation:

Precipitation is the creation of a solid from a solution. When the reaction occurs in a liquid solution, the solid formed is called the precipitate.

The formation of a precipitate indicates the occurrence of a chemical reaction.

Precipitation of carbide requires alot of energy which the diffusion activational energy alone cannot achieve and this was calculated to be 225.6 kJ/mol.

3 0
4 years ago
2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to
Korolek [52]

Answer:

Work input =283.47 KJ

Explanation:

Given that

P_1=150\ KPa

P_2=600\ KPa

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process  

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=mRT\ln \dfrac{P_1}{P_2}

So now putting the values

W=mRT\ln \dfrac{P_1}{P_2}

W=2.5\times 0.287\times 285\ln \dfrac{150}{600}

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

8 0
3 years ago
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