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Vera_Pavlovna [14]
3 years ago
12

Which of the following is true about a planet orbiting a star in uniform circular motion? A. The direction of the velocity vecto

r is always changing. B. The velocity of the planet is constant. C. The speed of the planet is always changing. D. The acceleration vector of the planet points directly away from the star.
Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
6 0

Answer;

A.The direction of the velocity vector is always changing.

Explanation;

-A circular orbit of a planet around a star is an example of uniform circular motion (its speed remains constant).

--When a planet has just the right speed (for a given radius of orbit), then it will travel in circular motion with a constant speed. This is called uniform circular motion. Note that the velocity vector changes direction although the speed is constant. This means that there must be a net force on the planet.

Luda [366]3 years ago
4 0
<span>As it is uniform circular motion therefore speed is constant. Therefore we can rule out option B. Also in circular motion the direction of velocity vector changes therefore velocity can't be constant. Therefore option B is incorrect as well. Also centripetal acceleration is always towards the center so option D is wrong as well. That implies option A is correct.</span>
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Answer:

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It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

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           Voy = (Y + ½ g t²) / t

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We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

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This is the time it also takes to travel the horizontal and vertical distance

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            H = Y = - ½ 9.8 2.83 2

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Answer:

1. v=14.2259\ m.s^{-1}

2. F_T=25.8924\ N

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Given:

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1.

\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel

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<em>Now, we find the linear mass density of the slinky.</em>

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30.7pC/εo = 3.47 V∙m <----- C)

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