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Juliette [100K]
3 years ago
9

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

t is the asteroid's orbital radius?
Physics
1 answer:
anzhelika [568]3 years ago
6 0

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
= 6.23 x 10^11m

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Vedmedyk [2.9K]

Answer:

The answer is a 11.25m/s

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3 years ago
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Why is the absolute magnitude of some stars greater than their apparent magnitude
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4 years ago
Two satellites revolve around the Earth. Satellite A has mass mmm and has an orbit of radius rrr. Satellite B has mass 6m6m and
True [87]

Answer:

r(b) = 2.45r

Explanation:

Parameters given:

Satellite A

Mass = m

Orbit radius = r

Satellite B

Mass = 6m

Orbit radius = r(b)

Gravitational force between Earth and Satellite A, F(A) is:

F(A) = (G*M*m)/r²

F(A) = GMm/r²

Gravitational force between Earth and Satellite B, F(B) is:

F(B) = (G*M*6m)/r(b)²

F(B) = 6GMm/r(b)²

Since F(A) = F(B),

GMm/r² = 6GMm/r(b)²

=> r(b)² = 6r²

r(b) = 2.45r

3 0
3 years ago
The estimated mass of the planet jupiter is 1.90 × 1027 kg and the density is believed to be 1.34 g/cm3. if jupiter were a perfe
V125BC [204]

We use the formula,

density=\frac{mass}{volume}

Given, mass=1.90\times 10^{27}\ kg = 1.90\times 10^{30}\ g and density =1.34\ g/cm^3.

Substituting these values, we get

volume = \frac{1.90\times 10^{30}\ g}{1.34\ g/cm^3} =1.4179\times 10^{30}\ cm^3.

As Jupiter were a perfect sphere, therefore the volume of sphere is given by

volume=\frac{4}{3} \pi r^3

Here, r is the radius of sphere.

Substituting the value of volume we get

1.4179\times 10^{30}\ cm^3=\frac{4}{3}\times 3.14\times r^3 \\\\ r^3=0.339\times 10^{30} \\\\r= 0.697\times 10^{10}\ cm.

The diameter is twice of radius, thus the diameter of Jupiter would be

2r=2\times 0.697\times 10^{10}\ cm=1.394\times 10^{10}\ cm

3 0
3 years ago
2. Do you expect the resistance of a light bulb to remain constant as the current through it is increased and the filament goes
koban [17]

Answer:

Remember that as the temperature of an object increases, the kinetic energy of the particles inside it also increases.

So, if the temperature of a wire increases, the kinetic energy of the particles inside increases. Then when the electrons try to flow through the wire, the probability of a collision is increased (then the resistance increases). Thus, if the filament goes from red-hot to white-hot (so the temperature of the wire increases) we can conclude that the resistance that the current experiences also increases.

So no, we can not expect the resistance of the light bulb to remain constant as the filament goes from red-hot to white-hot.

3 0
3 years ago
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