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Oxana [17]
2 years ago
9

Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The sec

ondary coil of this transformer is being replaced so that its output can be 730 kV for more efficient cross-country transmission on upgraded transmission lines.
Randomized Variables

Vi = 315 kV

V2 = 730 kV


(a) What is the ratio of turns in the new secondary to the number of turns in the old secondary?

(b) What is the ratio of new current output to the old current output for the same power input to the transformer?
Physics
1 answer:
SpyIntel [72]2 years ago
7 0

Answer:

  • 2.32
  • 0.43

Explanation:

12.0 kv primary voltage

315 kv secondary voltage ( converted voltage ) V1 or Vo

v2 (Vn)= 730 kv new secondary voltage

a) Ratio of turns in 730 kv to turns in 315 kv

\frac{Vn}{Vo} = \frac{Nn}{No} = \frac{730}{315}  therefore the ratio of turns = 2.317 ≈ 2.32

B) ratio of the new current output to the old current output for the same power input to the transformer

since the power input is the same

\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }     equation 1

Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current

therefore equation 1 becomes

\frac{In}{Io}  = \frac{Vo}{Vn} =  315 / 730 = 0.43

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3 years ago
Read 2 more answers
What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?
Artist 52 [7]

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

U=k\frac{q_1q_2}{r}    (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J

Hence, the energy between charges is 0.413 J

3 0
2 years ago
A motorcyclist drove 7 km at 57km/h and then another 7 km at 81 km/h. What was the average speed? ​
alex41 [277]

<u>Answer:</u>

<em>The average speed of the car is 66.9 km/h</em>

<u>Explanation:</u>

Here distance covered with the speed <em>57 km/h=7 km  </em>

distance covered with the speed of <em>81 km/h=7 km</em>

<em>Average speed is equal to the ratio of total distance to the total time. </em>

<em>total distance= 7 + 7= 14 km  </em>

<em>time= \frac{distance}{speed} </em>

<em>time taken to cover the first 7 km= 7/57 h  </em>

<em>time taken to cover the second part of the journey = 7/81 h </em>

<em>average speed =  14/(7/57+7/81)=(14 \times 57 \times 81)/945=66.9 km/h</em>

<u><em>Shortcut: </em></u>

<em>When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases. </em>

7 0
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Are outer planets gaseous
bija089 [108]

Answer:

Yes

Explanation:

The four outer planets are all gas giants made primarily of hydrogen and helium. They have thick gaseous outer layers and liquid interiors.

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2 years ago
Which planet moves the fastest and which planet moves slowest? a, b, c, d, e, or f. Explain
Alisiya [41]
D is the point where the planet moves the fastest. This is because it is in the perihelion, where the planet is moving at it’s fastest pace
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