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lbvjy [14]
2 years ago
7

Please help 25 points

Physics
1 answer:
luda_lava [24]2 years ago
3 0

This is a series of analysis of Angular velocity as relates to an asteroid's orbit. See the explanation below.

<h3>What is an asteroid?</h3>

Asteroids are stony bodies that circle the Sun.

Although asteroids circle the Sun in the same way as planets do, they are far smaller.

Hence, from the information given:

A) The square of the period is proportional to the cube of the semimajor axis, according to Kepler's third law. As a result, the period of Y equals (E) the period of Z.

B) Angular momentum is preserved here, hence it is equivalent (E).

C) As eccentricity increases, so does the angular momentum. In this case, Y and Z have the same period, and both satellites cover the same proportion of the territory in the same length of time.

This indicates that a satellite on Z must cover a lesser area in a given period of time than a satellite on Y. The area swept is approximately 1/2 the radius times the tangential displacement.

Because both satellites have the same "radius" at point y, the satellite on Z must have a lower tangential velocity than the one on Y. As a result, Y has more angular momentum than (G) Z.

D) Using Kepler's third law, X's period is bigger than (G) Z's.

E) In a circular orbit, the angular velocity is constant. As a result, the angular velocity of Y at y equals (E) that at s.

F) Z's angular velocity at c is smaller than (L) at i.

G) Y's angular velocity at y is larger than (G) Z's at y.

brainly.com/question/6860269

Learn more about Angular Velocity:
#SPJ1

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4 years ago
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Betelgeuse is 100,000 times more luminous than our sun, which means that it releases an estimated 3.846 x 1031 W of luminous lig
denis-greek [22]

Answer:

5.4  × 10⁸ W/m²

Explanation:

Given that:

The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W

the mass of the exoplanet = 5.972 × 10²⁴ kg

radius of the earth = 1.27 × 10⁷ m

half the distance (i.e radius r ) = 7.5  × 10¹⁰ m

a) What is the intensity of Betelgeuse at the "earth’s" surface?

The Intensity of  Betelgeuse  can be determined by using the formula:

Intensity \ I = \frac{P}{4 \pi r^2}

I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}

I = 544097698.8 W/m²

I = 5.4  × 10⁸ W/m²

8 0
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Hugs 5 things your dog might do when hugged
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2. li.ck: the dog will definitely li.ck you, as this is what dogs do when approached
3. pant: dogs pant all the time, especially if it's hot outside, so hugging them makes no difference
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How many revolutions per minute would a 25 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the
kozerog [31]
Refer to the diagram shown below.

Let ω =  the angular velocity (rad/s) of the wheel.

At the topmost point, the passenger, with mass m,  will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.

Therefore
mω²r = mg
ω = √(g/r)

Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s

Because 1 revolution per second is 2π rad/s, therefore
\omega = (0.8854 \,  \frac{rad}{s} )*( \frac{1}{2 \pi} \,  \frac{rev}{rad} )*(60 \,  \frac{s}{min} ) = 8.455 \, rev/min

Answer: 8.455 rev/min

8 0
3 years ago
Use this formula to solve this problem: Power (P) = Work (W) ÷ time (t) Peter's body supplies a force of 500 N to run up a 5 met
sesenic [268]
As you said p=w/t.
but, w=f×s
=500×5=2500j
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p=w/t
=2500/10=250 watts
4 0
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