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jek_recluse [69]

3 years ago

7

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of th

e force exerted on Q by q

1 answer:

Anon25 [30]3 years ago

8 0

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

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A manganese atom is pictured below.

Rufina [12.5K]

Manganese has 2 (two) electron that would free floating and able to form a metallic bond.
The electronic configuration of manganese is (Ar) 3d5 4s2. The two electron in 4s orbital are the valence electron which can freely move from one place to another.

5 0

3 years ago

Read 2 more answers

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done

Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

$W = F*d$

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

$F = W_{x} = mgsin(\theta)$

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²

θ: is the degree angle to the horizontal = 30°

$F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N$

Now, we can find the work:

$W = F*d = 24.53 N*20 m = 490.6 J$

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0

3 years ago

A second-grade teacher dropped a box of paper clips and they scattered all over the floor. She then asked her students, "Why wil

Yuliya22 [10]

The magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips.

<h3>Attraction of magnets</h3>

The like poles of magnets repel while unlike poles of magnets attracts. Magnets attracts irons or metallic materials.

The paper clips are mettalic or made of iron and hence the magnet will attract them.

There we can conclude that the magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips and will help to gather them together for easy picking.

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8 0

2 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

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

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A .100 kg bullet is flying at 200 m/s. How much energy is the bullet storing due to its motion?

morpeh [17]

KE = 2000 J

Explanation:

KE = (1/2)mv^2

= (1/2)(0.100 kg)(200 m/s)^2

= 2000 J

4 0

2 years ago