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3 years ago

Why do electric field lines explain why like charges repel and opposite charges attract?

1 answer:

5 0

It can be explained as follows: consider the field produced by a positive charge. If we place a positive test charge in this a field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel)

Explanation:

Electric fields are vector fields, and they are represented using field lines.

The field lines give indications on both the magnitude and the direction of the electric field. In fact:

The magnitude of the field can be inferred from the spacing between the lines: the closer the lines are, the stronger the field, while for a weaker field the lines are more spread apart
The direction of the field is given by the direction of the field lines

In particular, by convention the direction of the field lines represent the direction of the force that a positive test charge would feel when immersed in that field: this means that a positive test charge would accelerate in the direction of the field lines, while a negative test charge would accelerate in the direction opposite to the field lines.

This is in agreement with the fact that like charges repel and opposite charges attract. In fact, the lines of the electric field produced by a single-point positive charge point away from the positive charge: if we place a positive test charge in this field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel).

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A book prone to air resistance is released from rest 300 m

yaroslaw [1]

Answer:

Approximately $73\%$ .

(Assuming that $g = \rm 9.81\; m \cdot s^{-2}$ .)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let $m(\text{book})$ represent the mass of the book.

The initial potential energy of the book is

$\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}$ .

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be $(2.943\times 10^3) \cdot m(\text{book})$ .

When the book was about to hit the ground, its speed is $\rm 40\; m \cdot s^{-1}$ . Its kinetic energy would be:

$\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}$ .

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be $\left(8.00\times 10^2\right)\cdot m(\text{book})$ when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

$\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}$ .

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, $m(\text{book})$ , was eliminated in this process.

$\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}$ .

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4 years ago

A sheet of paper is about 0.0970 mm thick. what is the volume (in cubic inches) of a standard 8.50 in.×11.0 in. sheet of paper?

umka21 [38]

height = .0970 mm = .00831 inches

Volume = length * breadth * height = 8.5 * 11 * 0.00381

Volume = 0.776985 inches^cube

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3 years ago

A meter stick is place in a very high-speed spaceship. What length would the astronauts say the meter stick was? What would the

Marysya12 [62]

Answer:

a) The astronauts would see the real length of the meter stick, i.e. L₀

b) The length of the meter stick as measured by the stationary observer will be $L = L_{0} }{\sqrt{(1-(\frac{v}{c} )^{2} } }$

Explanation:

a) Let the proper length of the meter stick be L₀

The meter stick and the astronauts on the on the space ship are on the same moving frame, therefore, they will see the exact length of the meter stick, that is, L₀

b) A stationary observer watching the space ship and meter stick travel past them will see a contracted length of the meter stick

The original length = L₀

Let the speed of the space ship = v

The contracted length, L, is related to the original length in the frame of rest by

L = L₀/γ......................(1)

Where γ = $\frac{1}{\sqrt{(1-(\frac{v}{c} )^{2} } }$ ....................(2)

Substituting equation (2) into (1)

$L = L_{0} }{\sqrt{(1-(\frac{v}{c} )^{2} } }$

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3 years ago

Two observers in different inertial reference frames moving relative to each other at nearly the speed of light see the same two

lions [1.4K]

Answer:

The correct answer is d Both the observer's are correct

Explanation:

We know by postulates of relativity that laws of physics are same in different inertial frames.

Thus for each of the frames they make observations related to their frames and since the observations are true for their individual frames they both are correct. But when we compare the two frames we need to use transformation equations to compare both the results.

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3 years ago

Need help ASAP

devlian [24]

Answer:

cell pohnes convert sound waves into radio waves

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To send out a radio signal far and wide.. it is called: broadcast

Explanation:

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3 years ago

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