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Lelechka [254]
3 years ago
14

Two long straight wires are suspended vertically. The wires are connected in series, and a current from a battery is maintained

in them. What happens to the wires? What happens if the battery is replaced by an a-c source?
Physics
2 answers:
GREYUIT [131]3 years ago
8 0

Answer:

(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.

(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.

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GuDViN [60]3 years ago
4 0

Explanation:

<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>helps</u></em><em><u> </u></em><em><u>you</u></em><em><u> </u></em><em><u>lot</u></em><em><u>!</u></em>

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4 years ago
Which best explains parrallel forces
svetlana [45]

Answer:

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8 0
3 years ago
Jan needs 10 meters of yarn. She has 746 centimeters. How many more cm does she need​
MAVERICK [17]

Answer:

254

Explanation:

100 centimeters is equal to 1 meter.

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6 0
3 years ago
Read 2 more answers
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

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