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babymother [125]

2 years ago

14

12 inches is equivalent to_ foot.A. 1B. 12C. 24D. 36

1 answer:

Tamiku [17]2 years ago

6 0

Answer:

A. 1

Explanation:

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Name the seven physical qualities for which standards have been developed.

SIZIF [17.4K]

Answer:

HUMAN DEVELOPMENT

MOTOR BEHAVIOR

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MEASUREMENT AND EVALUATION

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Explanation:

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4 years ago

An automotive Battery is rated at120 A-h. This means that under certain test conditions, it canoutput 1 A at 12 V for 120 hours.

Hitman42 [59]

Answer:

Part A:

In W-h:

Energy Stored=1440 W-h

In Joules:

$Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ$

Part B:

In W-h:

Energy left=240 W-h

In Joules:

Energy left= 8.64*10^5 J

Explanation:

Part A:

We are given rating 120A-h and voltage 12 V

Energy Stored= Rating*Voltage (Gives us units W-h)

Energy Stored=120A-h*12V

Energy Stored=1440 W-h

Converting it into joules (watt=joules/sec)

Energy Stored= $1440 Joules * \frac{3600sec}{h}h$

Energy Stored=5184000 Joules

$Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ$

Part B:

Energy used by lights for 8h=150*8

Energy used by lights for 8h=1200W-h

Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h

Energy left=1440-1200

Energy left=240 W-h

Energy left= $240 Joules * \frac{3600sec}{h}h$

Energy left=864000 Joules

Energy left= 8.64*10^5 J

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3 years ago

A cold-rolled sheet metal that is 40 mm wide and has a thickness of 5.00 mm is going to be bent into a V shape with a 60° angle.

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a= the force of gravity b= the amount of bicker to maple syrup ratio

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3 years ago

Which of the following are made up of electrical probes and connectors?

Mariulka [41]

Uhm is there a multiple choice?

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