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3 years ago

What’s true about the elliptical path that the planets follow around the sun?

Physics

2 answers:

lukranit [14]3 years ago

8 0

Answer:

C. A line can be drawn from the planet to the sun that sweeps out equal areas in equal time

Explanation:

From Kepler's law of planetary motion we know that planets in the solar system revolve in elliptical orbits around the sun. The line which joins the planet and the sun sweeps equal areas in equal interval. As a consequence of this, the planets which are closer to sun have smaller period of revolution than the planets which are farther away.

madam [21]3 years ago

3 0

<h2>Answer: A line can be drawn from the planet to the Sun that sweeps out equal areas in equal times</h2>

Kepler’s Laws are three mathematic laws to describe the<u> movement of the planets around the Sun</u>, but it can be generalized for the <u>movement of any body orbiting a bigger one</u>, for example, The Moon orbiting the Earth.

Now, according to the Second Kepler’s Law of Planetary motion:

<em>In equal times, the areas swept by the planet in its orbit around the Sun are equal. </em>

For this to be possible, the speed of the planet must vary. Hence, the planet will move rapidly near the Sun (perihelion) and move slowly when it is away from the Sun (aphelion).

Therefore the correct option is C.

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Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion. Choose the planetesimal t

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Answer:2

Explanation:

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4 years ago

A bag of wheat 60kg find the height to which it is lifted so that its potential energy is 3000j

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PE = mgh
m = 60 kg
g = 10 m/s
h = X
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3000 = 60×10×X = 5 m

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4 years ago

Which kind of information is known about a person if that person’s DNA is in a database? A. how old the person is

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3 years ago

Read 2 more answers

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

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4 0

2 years ago