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AfilCa [17]
3 years ago
5

How does the force exerted by two magnets change as the magnets are moved farther apart?

Physics
1 answer:
galben [10]3 years ago
8 0
<span>The magnetic force exerted by two magnets decreases as the magnets are moved farther apart.

</span><span>The magnetic force between two magnets is the force of repulsion or attraction between two magnets which leads to the calculation of a positive work.
</span><span>The magnetic of a magnet is strongest at its poles.</span>
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A sample contains 16 g of a radioactive isotope. how much radioactive
mafiozo [28]

After one half-life, 8 g of radioactive isotope will remain in the sample.

<h3>What is radioactivity?</h3>

The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

The following formula is used to compute the number of half lives elapsed:

\rm N=\frac{N_0}{2^n} \\\\ N=\frac{16}{2} \\\\ N= 8 \ gram

Hence,8 gram of radioactive isotope remains in the sample after 1 half-life.

To learn more about the radioactivity, refer  to the link;

brainly.com/question/1770619

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7 0
2 years ago
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Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad
Xelga [282]
0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is  1.4129 x 10⁻⁸   the radius of the Earth.
6 0
3 years ago
A negatively charged rubber rod can pick up small, neutrally charged pieces of paper by a process called
Ronch [10]
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4 0
3 years ago
A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
3 years ago
IF YOU ANSWER THESE 4 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!
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Answer:

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