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Troyanec [42]
3 years ago
9

A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutr

al. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________
Physics
1 answer:
stealth61 [152]3 years ago
4 0

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the   positive charge in the piece of paper.

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A straight railroad track is being built to connect two cities. the measured distance of the track between the two cities is 160
elena55 [62]
Let point A be 0.0 miles (first city)
Let point B be 160.5 miles (first city to second city)
Let point C be 28.5 miles (first city to mail stop)

Take C – A = C [28.5 - 0.0 = 28.5] (This checks the distance between city 1 and Mail stop) 

Then Take B – C = Distance from the first city to the second city [160.5 - 28.5 = 132 Miles]


Answer: The Mail stop is 132 miles from the Second City.
7 0
3 years ago
The scientific method can easily be misinterpreted as ___________. Question 1 options: a great magical trick, or slide of hands,
evablogger [386]

Answer:

I think the answer is  

a "cookbook" recipe for performing scientific investigations

Explanation:

5 0
3 years ago
Read 2 more answers
A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pus
Alchen [17]

Answer:

<h3>2,321.62Joules</h3>

Explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone  = 2,321.62Joules

<em>Hence the amount of work done is 2,321.62Joules</em>

3 0
3 years ago
What is specific gravity​
Mrac [35]

Answer:

It is the ratio of the density of a substance to the density of a given reference material.

Explanation:

<em>Specific gravity is also known as relative density.</em>

<u>To find the relative density of substance, you:</u>

  • Divide the density of substance measured
  • And divide that by the density of the reference

7 0
2 years ago
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

4 0
3 years ago
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