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3 years ago

A positively charged light metal ball is suspended between two oppositely charged metal plates on

Physics

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

The positively charged ball moves between both charged plates till the plates and the ball all become neutral.

Check Explanation for more.

Explanation:

Let the ball be in square brackets, and the plates in normal brackets.

(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.

The ball hanging on the insulated thread becomes neutral too at this point.

Hope this Helps!!!

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An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce

ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

From them we can get $v^2=v_0^2+2ad$ .

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}$

We multiply both sides by 2a, and continue:

$2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Being d the displacement $x-x_0$ , we have $v^2=v_0^2+2ad$

For our exercise, we will write this as:

$d=\frac{v^2-v_0^2}{2a}$

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

$d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m$

For the time we use:

$t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s$

6 0

3 years ago

How to find velocity

Valentin [98]

V=d/t

v=speed
d=distance traveled
t=time taken

hope this helps

8 0

3 years ago

Find the acceleration of a car with the mass of 1,200 kg and a force of

Mashutka [201]

Answer:9.17 m/s^2

Explanation:

mass=1200kg

Force=11 x 10^3 N

Acceleration=force ➗ mass

Acceleration=11 x 10^3 ➗ 1200

Acceleration=9.17

Acceleration=9.17 m/s^2

3 0

3 years ago

Can someone please help with this problem?

andriy [413]

no question please so what's the problem

4 0

3 years ago

In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of

nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

$F_g = F_c$

$\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}$

now we can say that kinetic energy of satellite is given as

$KE = \frac{1}{2}M_s v^2$

$KE = \frac{GM_sM_p}{2r}$

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

$U = -\frac{GM_sM_p}{r}$

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0

3 years ago