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3 years ago

What do electric power companies deliver to customers

Physics

1 answer:

kvv77 [185]3 years ago

8 0

Answer:

Power plants generate electricity that is delivered to customers through transmission and distribution power lines. High-voltage transmission lines, such as those that hang between tall metal towers, carry electricity over long distances to meet customer needs.

HOPE THIS HELPED! XDDDDDDDDD

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What is the necessary conditions for the production of sound?

TEA [102]

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Answer: Something that's vibrating, and you also need medium for those vibrations to start in.

I hope this helped!

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8 0

2 years ago

A raging bull of mass 700 kg runs at 36 km/h .How much kinetic energy does it have ?

Ipatiy [6.2K]

Answer:

Use equation for kinetic energy: Ek=mV²/2

m=700 kg

V=10m/s

Ek=700kg*100m²7s²/2

Ek=35000 J=35kJ

Explanation:

Hope this helps you

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5 0

3 years ago

Read 2 more answers

An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g 10 . If it has

RoseWind [281]

Answer:

Work done against gravity will be

W = Mgℓ

Explanation:

Work done to raise the mass from ground to given height is against gravity

So here work done is given by the formula

$W = F.d$

here we know that

$F = Mg$

it is the force due to gravity which is also known as weight

so here distance moved by the object is given as

d = ℓ

now work done is given as

W = Mg ℓ

8 0

3 years ago

Read 2 more answers

How do electric and magnetic fields interact in an electromagnetic wave?

kotegsom [21]

Answer:

Electric and magnetic field waves are oriented at 90 degree angles relative to each other.

Explanation:

8 0

3 years ago

The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for

abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

$\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}$

5 0

3 years ago