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3 years ago

The mantle in the earths core continuously heat up and cool down using convection currents

Physics

2 answers:

o-na [289]3 years ago

7 0

This statements is true

Nostrana [21]3 years ago

3 0

Yes, The statement is True

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Complete this equation that represents the process of nuclear fission.

tekilochka [14]

Answer:

A: 146

B: 56

Explanation:

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3 years ago

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an object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. the object strikes the ground 3.

Olegator [25]

You know that the formula for finding horizontal displacement is;

s=uₓt
s=(5 m/s)(3s)
s=15 m

Therefore, the ball travelled 15 m.

Hope I helped :)

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3 years ago

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which of the following is correct for solids? a. y=2/3 alpha b.y= 2/3 bita c.y = 3/2 bita d.y=3/2 alpha

8_murik_8 [283]

option a ...

please mark hcfhvx

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3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

A system consists of N particles that can occupy two energy levels: a nondegenerate ground state and a three-fold degenerate exc

cupoosta [38]

Answer:

Ng = 0.893 N, Ne = 0.107N

Explanation:

Number of particles in Ground state = Ng

Number of particles in Excited state = Ne

Ne/Ng = e^{(-ΔE)/kt}

Since excited state is 3 fold degenerate

Ne/Ng =3 x e^{(-ΔE)/kt}

ΔE = Energy difference between ground and excited states = 0.25eV

T = 960 K

Constant k = 8.617 x 10^-5 eV/K

Ne/Ng = 3 x e^{-0.25/(8.617x10^-5) x 960}

= 3 x e^(-3.188645)

= 3 x 0.0412 = 0.1237 ≅ 0.12

Ne = 0.12 Ng

but Ne + Ng = N, where is N is total number of particles, substituting Ne into equation we get,

Ng(1 + 0.12) = N

Ng = N/1.12 = 0.893N

and Ne = 0.12 x 0.893 N = 0.107 N

3 0

3 years ago