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Lady_Fox [76]
2 years ago
12

For long-distance transmission of power, in order to reduce losses in the wires by driving a low current at high voltage, we sho

uld
a) use a step-up transformer

b) change the length of the wires between transmission towers

c) not use a transformer at all

d) change the type of conducting material

e) use a step down transformer
Physics
2 answers:
rewona [7]2 years ago
6 0

ans d) change the type of conducting matey

Snezhnost [94]2 years ago
3 0

For long-distance transmission of power, in order to reduce losses in the wires by driving a low current at high voltage, we should use a step-up transformer (a) .

(That's at the beginning, to step-up the voltage for long-distance transmission.  And then use a step-down transformer at the end, to reduce the voltage just before we pour it into the customer's house.)

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Which characteristics describe a point charge?
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Answer:

C

Explanation:

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3 years ago
You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di
kherson [118]

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

  • 100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

  • 1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

  • 116 km/1 h = 800 km/? h
  • 800 = 116 · ?
  • ? = 800/116
  • ? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

  • 12 h - 6.89655172 h = 5.10344828 h  

Using this remaining time and my remaining distance, I can calculate my average speed.

  • ? km/1 hr = 400 km/5.10344828 h
  • 5.10344828 · ? = 400
  • ? = 400/5.10344828
  • ? = 78.3783783148  

My average speed during the last third of the distance is around 78 km/h.

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2 years ago
A young boy is dragging his 10 kg stuffed animal by the tail at a constant velocity of 4 m/s. The coefficient of friction betwee
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Explanation:

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2 years ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
NemiM [27]

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

5 0
2 years ago
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