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Anna11 [10]
3 years ago
9

A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

awhile and finally runs quickly back to her initial position along a straight line.
Required:
Draw displacement versus time plots best represents the student’s trip?
Physics
1 answer:
True [87]3 years ago
6 0

Answer:

The first interval is walked slowly, this is a straight line with a small slope

Second interval stops, which gives a horizontal line, indicating the same position

Third interval, walk back, straight downhill

Explanation:

In this problem we have a uniform movement, this means that the acceleration in each intervals

              x = v t

 

The first interval is walked slowly, this is a straight line with a small slope

Second interval stops, which gives a horizontal line, indicating the same position

Third interval, walk back, straight downhill

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Find electric field at point p which is a distance l away from the both +q and -q
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Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

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