Physics, 31.12.2019 05:31 lexybellx3

Apositive point charge q is fixed on a very large horizontal frictionless tabletop. a second positive point charge q is released from rest near the stationary charge and is free to move. which statement best describes the motion of q after it is released? as it moves farther and farther from q, its speed will keep increasing. as it moves farther and farther from q, its speed will decrease. as it moves farther and farther from q, its acceleration will keep increasing. its speed will be greatest just after it is released. its acceleration is zero just after it is released. a charge q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. at a point p that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet e. if the sheet is now stretched so that its sides have length 2d, what is the magnitude of the electric field at p? e e/2 2e e/4 4e

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Answer from: nathaniel12

Explanation:

When a second positive point charge q is released from rest near the stationary charge and is free to move , the stationary charge will exert a repulsive force upon it which will go on decreasing as the mobile charge moves away from stationary charge. So it will have decreasing but positive acceleration . So its velocity will go on increasing but with decreasing rate.

So the correct statement is

"As it moves farther and farther from Q, its speed will keep increasing".

The electric field near a charged sheet is proportional to the charge density on the charged surface .As the area has been increased 4 times , area charge density will become 1/4 times , so electric field near it will also reduce to 1/4 times . Hence new electric field will be

E / 4 . Ans .

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

The sun's gravitational pull is weaker on the planets that are closer to it, i think.

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