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frutty [35]
3 years ago
9

A(n) __________ is a gravitationally curved path that a celestial body travels around a point in space. A(n) __________ is a gra

vitationally curved path that a celestial body travels around a point in space. retrograde motion small circle epicycle planet orbit
Physics
1 answer:
joja [24]3 years ago
5 0

Answer: A planet orbit is a gravitationally curved path that a celestial body travels around a point in space.

Explanation:

We define planet orbit as the path that a planet (or any other celestial object) has around two focal points (the orbit is an ellipse, the ellipses have two focal points). For example, the moon's orbit is around the Earth is almost circular, while Earth orbit is less circular. This movement is caused by the strong gravitational attraction of the sun, that keeps all the planets fixed in the orbits, this is why this is called a "gravitationally curved path".

the correct option is planet orbit.

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A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr
MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

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3 years ago
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Suppose that a spiral galaxy is located at the center of a spherically symmetric dark matter halo
Novosadov [1.4K]

To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.

In general terms the Orbital speed is described as,

V_{orbit} = \sqrt{\frac{G\rho 4\pi r^3}{3}}

PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,

\frac{4\pi G\rho r}{3} = \frac{v^2}{r}

\frac{4\pi G\rho r}{1} = \frac{3v^2}{r}

\frac{\rho}{1} = \frac{3v^2}{r^2 4\pi G}

\rho = \frac{1}{r^2}

PART B) This time we havev=\omega t, where \omega is the angular velocity (constant at this case)

\frac{4\pi G\rho r}{3} = \frac{v^2}{r}

\frac{4\pi G\rho r}{3} = \frac{(\omega r)^2}{r}

\rho = \frac{3\omega r}{4\pi Gr}

\rho = \frac{3\omega^2}{4\pi G} \propto constant

PART C) If the total mass interior to any radius r is a constant,

\frac{4\pi G\rho r}{3} = \frac{v^2}{r}

\frac{GM}{r^2}=\frac{v^2}{r}

v = \sqrt{\frac{GM}{r}}

v= \sqrt{\frac{1}{r}}

3 0
4 years ago
Determine the value of the force exerted by the surface (normal force) on a
Advocard [28]

Answer:

53.5 N

Explanation:

Vertical component of the F force   50 sin30    = 25 N  upward

   force of gravity = m g = 8 * 9.81 =78.5 N Downward

NET downward force by block on table = net upward force exerted by table =  78.5 -25 =53.5 N

8 0
2 years ago
A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 10.0 min at 50.0
irakobra [83]
<h2>The average speed for the entire trip is 47.5 m/s .</h2>

It is given that for different time span car have different speed and also the person spend 40\ min=\dfrac{40}{60}\ hrs =0.67\ hrs\ . in eating lunch and buying gas.

We know , average speed is total distance covered by total time taken .

Therefore , average speed , v=\dfrac{total\ distance }{total\ times}

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Hence, this is the required solution.

Learn More :

Average speed

https://brainly.in/question/12701198

7 0
3 years ago
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