Ask question

Ask question

All categories

3 years ago

12

Two similar gold spheres are separated by a distance of

1 answer:

solong [7]3 years ago

5 0

Answer:

The answer to your question is the letter D. 2.5 N

Explanation:

The electrostatic force is the same in both directions,

If the electrostatic force on B due to A is 2.5 N, the magnitude of the electrostatic force on A due to B must be 2.5N.

Maybe the direction is different but the magnitude is the same.

You might be interested in

What color is mercury

Evgen [1.6K]

Hello!
If you are referring to the mercury substance, mercury is a bright red color, usually used in early thermometers and is deadly if humans have a lot of constant with this substance

7 0

3 years ago

Read 2 more answers

A standing wave is produced by plucking a string. the points along the plucked string that appear not to be vibrating are produc

Rasek [7]

The answer is destructive

4 0

3 years ago

Read 2 more answers

In which of the following situations are there a force on the object balanced

ser-zykov [4K]

Where’s the rest at????

3 0

3 years ago

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

sweet [91]

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

Where k= $9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

3 0

3 years ago

a 4.0 kilogram ball moving at 8.0 m/s to the right collides with a 1.0 kilogram ball at rest. after the collision the 4.0 kilogr

Inga [223]

Step by step equation

7 0

3 years ago