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wlad13 [49]
3 years ago
8

What are brief explanation of charging by a conduction and charging by Induction​

Physics
1 answer:
Lina20 [59]3 years ago
5 0

Charging by conduction involves the contact of a charged object to a neutral object.

Induction is where the charged object is brought near but never contacted to the object being charged.

I saw your recent post, I never saw these in my feed.

I hope this helps you fam :)

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The first stage in the GAS model of stress is
Vladimir [108]
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8 0
4 years ago
How can i calculate distances between objects by using the concepts of echo location
dangina [55]
Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return
5 0
3 years ago
A car speed off around a bend at a constant 10m/s explain why it's velocity is not constant
VARVARA [1.3K]
It's velocity is not constant as direction is changing.
 
We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.

Hope this helps!
8 0
3 years ago
the first s-wave reaches a seismic station 22 minutes after an earthquake occurred. how long did it take the first p-wave to rea
Naddik [55]

The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

<h3>Time of travel of the P-wave</h3>

In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.

<h3>Relationship between speed and time</h3>

v ∝ 1/t

v₁t₁ = v₂t₂

t₁/t₂ = v₂/v₁

t₁/t₂ = 0.6v₁/v₁

t₁/t₂ =  0.6

t₁ = 0.6t₂

t₁ = 0.6 x 22 mins

t₁ = 13.2 mins

Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

Learn more about P-waves here: brainly.com/question/2552909

#SPJ1

7 0
2 years ago
A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d
Minchanka [31]

This problems a perfect application for this acceleration formula:

         Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:     1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half:    1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by  0.65 m/s²: 

                         T² = (1600 m) / (0.65 m/s²)

                         T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

                         T =  2 √(1600/0.65) = <em>99.2 seconds</em>  (rounded)


7 0
3 years ago
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