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DiKsa [7]
4 years ago
9

Can charging by friction occur only in solids? explain using an example.

Physics
1 answer:
frozen [14]4 years ago
8 0
The answer to the question can charging by friction occur only in solids is no. not at all. the static charge generated by following fluids can be hazardous. just as the static discharge when you touch a doorknob produces a spark, lightening is the spark from the static charge built up in the atmosphere.
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Gravity is the main force for mass movement.<br> True<br> False
Travka [436]

Hello!

Mass, a weight.

Gravity, force.

Translated the Turkish.

Kütle, bir ağırlıktır.

Yer çekimi, insanın yere uyguladığı kuvvet.

Answer (Turkish : Cevap)

True (Doğru)

Have a nice day! ( İyi günler!)

5 0
3 years ago
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What quantity do units represent in a value?
mihalych1998 [28]

Answer : Magnitude

Explanation :

In a value, the magnitude is represented by its units. It can be adopted by convention or by law.

Magnitude of any unit is used to measure the same kind of quantity.

For example: The unit of length which is a physical quantity is meter (m).

So, magnitude is correct answer.

5 0
3 years ago
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Compare and contrast electrical and magnetic forces
erik [133]

Answer:

the eletrical forces are created when an object is both moving charges and stationary charges.

the magnetic forces are created when an act on only moving charges.

3 0
3 years ago
Do longitude lines run horizontally (east-west) or vertically (north-south)?
nlexa [21]
Longitude- Horizontal (East West)
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3 0
3 years ago
An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
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