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2 years ago

8

How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another

?

1 answer:

Harman [31]2 years ago

4 0

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

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White visible light travels from the Sun to Earth. What is the best description of white light?

aliya0001 [1]

Answer:

As investigated by Sir Isaac Newton,the white light contains all colours of light.

This shorten by a term rainbow colours.

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2 years ago

Will mark as BRAINLIEST....... The Displacement x of particle moving in one dimension under the action of constant force is rela

pentagon [3]

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

$x=4t^3+3t^2-5t+2$

Where,

x is in meters and t is in sec

We know that,

Velocity,

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5$

(a) i. t = 2 s

$v=12(2)^2+6(2)-5=55\ m/s$

At t = 4 s

$v=12(4)^2+6(4)-5=211\ m/s$

(b) Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6$

Pu t = 3 s in the above equation

So,

$a=24(3)+6\\\\a=78\ m/s^2$

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3 years ago

A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,

vfiekz [6]

The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
$J=F \Delta t$
But the impulse is also equal to the change in momentum of the cannonball:
$J=\Delta p$
If we put the two equations together, we find
$F \Delta t= \Delta p$
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
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3 years ago

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$

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3 years ago

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2 years ago