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2 years ago

Determine the orbits period of the moon when the distance between the earth and the moon is 3.82 x 10 to the power of 8

Physics

1 answer:

alexandr402 [8]2 years ago

4 0

Answer

From

V=Distance/time

The distance round a circular path or Object is 2πr

v=2πr/T

Making T(period) subject

T=2πr/v

where v is the linear velocity.

We don't have the velocity but we can get it.

The Moon and Earth exert gravitational force on each other and the Moon is kept in Orbit by Centripetal force.

Since the Moon doesn't fly out of Orbit... it must mean that the Gravitational force being exerted on it by the Earth is equal to its centripetal Force.

Equating Both (Fg=Fc)

Fg=GMm/R²

Fc=mv²/r

GMm/R² = mv²/R

Canceling out "m" and "r" on both sides

we're left with

V²=GM/R

V=√GM/R

Where M= Mass of Earth (5.98x10^24)

R=Distance between the center of earth and the Moon

G= Gravitational Constant(Value 6.67x10^-11)

V=√6.67x10^-11 x 5.98x10^24/(3.82x10^8)

V=1021.84meters per second.

Now

T=2πr/v

=2π x 3.82x10^8 / 1021.84

T=2.35x10^6seconds

Converting to days by dividing by(24 x 3600)

You have

T=27.2days approx.

✅✅✅

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3 years ago

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A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

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3 years ago

A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field

arsen [322]

magnetic field due to a finite straight conductor is given by

$B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)$

here since it forms an equilateral triangle so we will have

$\theta_1 = \theta_2 = 60 degre$

also the perpendicular distance of the point from the wire is

$r = \frac{a}{2\sqrt3}$

now from the above equation magnetic field due to one wire is given by

$B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)$

$B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)$

$B = \frac{3\mu_0 i}{2\pi a}$

now since in equilateral triangle there are three such wires so net magnetic field will be

$B = \frac{9\mu_0 i}{2\pi a}$

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3 years ago

Which of the following best describe a generator ?

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3 years ago

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A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe

Simora [160]

Answer:

Final temperature is 295K

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Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C× $\frac{1mol}{63,546g}$ ×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C× $\frac{1mol}{18,02g}$ ×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK× $\frac{1mol}{63,546g}$ ×25,0g×(X-363K) = -75,2J/molK× $\frac{1mol}{18,02g}$ ×100,0g× (X-293K)

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I hope it helps!

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