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2 years ago

What is a dwarf planet?

2 answers:

6 0

Hey, Sverlnexlcasatyr!
A dwarf planet looks like a planet, but is missing things to become a planet. For example, Pluto is now a dwarf planet.
I hope this helps;)

Kamila [148]2 years ago

5 0

A dwarf planet is a small celestial body resembling a planet, but lacks necessary things to be a planet

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A north magnetic pole is facing another north magnetic pole with a distance x. If the distance between the poles becomes 12x, wh

krek1111 [17]

Answer:

1.)The field energy will increase.

The rest of the answers:

2.)The energy increases, and the lines of force are denser

3.) It points toward the field of earths magnetic poles.

4.) l, ll, and lll only

5.) ll, lV, l, lll

4 0

3 years ago

Highest density if electrostatic charge in a metal is found in

arsen [322]

There’s no picture so how r we supposed to answer it

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2 years ago

3. Neha travels 4 m towards the north, 3m towards the east, and then 7 m towards south, find the displacement and total distance

Digiron [165]

Answer:

Explanation:

Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.

7 0

3 years ago

Kinetic energy problemset

Sonja [21]

Answer:

KE = 4 mv2 m = 2xKE valami. V m.

Explanation:

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3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

6 0

2 years ago