What Is An Insulator Give An Example

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

4 0

4 years ago

First to answer will be the brainliest I need the answer ASAP don't answer if you don't know the answer

Phantasy [73]

Answer:

The answer is D you were correct

8 0

3 years ago

Read 2 more answers

Problem: At the local swimming hole, a favorite trick is torun

denis23 [38]

Answer:

2 revolutions

Explanation:

Assume that when she runs off the edge of the 8.3m high cliff, her vertical speed is 0. So gravitational acceleration g = 9.8m/s2 is the only thing that makes her fall down. So we can use the following equation of motion to calculate the time it takes for her to fall down:

$s = gt^2/2$

where s = 8.3 m is the distance that she falls, t is the time it takes to fall, which is what we are looking for

$t^2 = \frac{2s}{g} = \frac{2*8.3}{9.8} = 1.694$

$t = \sqrt{1.694} = 1.3 s$

Since she rotates with an average angular speed of 1.6rev/s. The number of revolutions she would make within 1.3s is

$rev = 1.3 * 1.6 = 2 revolution$

8 0

4 years ago

the smallest unit in physics is the Planet length.wr need to know constant (h) the speed of light (c) and Newton's Gravitational

fgiga [73]

Yes, that's right. It's the 'Planck' length, not the 'Planet' length.

You could easily find these with a web search. But in gratitude
for the bountiful 5 points, I've saved you the trouble.
AND guess what !   By doing that, I learned something, and
you didn't.

Speed of light (c):    299,792,458 meters per second

Gravitational constant (G): 6.67 x 10⁻¹¹ newton-meter²/kilogram²

Planck's Konstant (h):   6.63 x 10⁻³⁴ joule-second

Planck Length:    1.6 x 10⁻³⁵ meter
(about 10⁻²⁰ the size of a proton)

Planck Time:    10⁻⁴³ second
   (about the time it takes to travel
a Planck Length at the speed of light)

4 0

3 years ago

Read 2 more answers

The impulse formula allows us to relate the concepts of mass and velocity (an objects momentum) to that of:

NikAS [45]

Answer:

3: force acting on an object over time

Explanation:

The impulse formula is: change in momentum = force x time

Δp = f x t

3 0

3 years ago