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Ronch [10]
2 years ago
8

If the mass of the sun is 1x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____

, and ____, and all planets will orbit the sun successfully.
Physics
1 answer:
Minchanka [31]2 years ago
5 0

If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.

<h3>
What are planets?</h3>

Planets are the large spherical shaped objects that rotate about the Sun in the elliptical orbits.

Planets are shaped from Planetary cloud. The dust storm and gases gathers under its own weight. The dense matter beginnings pivoting at high paces and accumulates more mass. The center structures, the star and rest of it ultimately levels into a curved plate from which planet is formed.

Thus,  if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.

Learn more about planets.

brainly.com/question/14581221

#SPJ1

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A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child appl
lozanna [386]

Answer:

The value is KE = 259.6 \  J

Explanation:

From the question we are told that

     The weight of the horizontal solid disk is  W = 805 \  N

      The radius of the horizontal solid disk is  r =  1.58 \  m

      The force applied by the child is  F  =  49.5 \  N

       The time considered is  t =  2.95 \  s

Generally the mass of the  horizontal solid disk is mathematically represented as

          m_h  =  \frac{W}{ g}

=>       m_h  =  \frac{805}{ 9.8 }

=>       m_h  =  82.14 \  N

Generally the moment of inertia  of the horizontal solid disk is mathematically represented as  

         I  =  \frac{1}{2} *  m *  r^ 2

=>      I  =  \frac{1}{2} *  82.14 *   1.58^ 2  

=>      I  =  102.5 \  kg \cdot m^2

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

           T =  I  *  \alpha   =  F *  r

=>         \alpha  =  \frac{ F  *  r }{ I }

=>         \alpha  =  \frac{  49.5  *   1.58 }{  102.53 }

=>         \alpha  = 0.7628

Gnerally from kinematic equation we have that

         w =  w_o  +  \alpha t

Here  w_o is the initial angular velocity velocity of the horizontal solid disk  which is  w_o  =  0\   rad/s

So

           w =   0  +  0.7628 * 2.95

=>        w =  2.2503 \  rad/s

Generally the kinetic energy is mathematically represented as

        KE =  \frac{1}{2}  *  I  *  w^2

=>      KE =  \frac{1}{2}  * 102.53  *  2.2503 ^2

=>      KE = 259.6 \  J

8 0
3 years ago
What is the change in its velocity, v, during this 0.80-s interval?
lyudmila [28]
This link should help you out https://quizlet.com/40330134/biomechanics-questions-flash-cards/
7 0
3 years ago
Find the angle of prism if the ray just fail to emerge from 2nd face When ray of light falls normallly on the face of prism of R
Usimov [2.4K]

The angle of prism is 41.81 degrees.

<u>Explanation:</u>

For no emergence to be taken place, inside a prism, Total Internal Reflection (TIR) should take place at the second surface.  For TIR, at second surface, angle of refraction must be greater than critical angle. Angle of prism is related to refraction as,

                            A>r_{1}+C

Since, r_{1} = C and A \geq 2 C

This implies A \geq C

                         \sin A \geq \sin C

                         \sin A \geq \frac{1}{\mu}

                         \sin A \geq \frac{2}{3}

when sin goes to other side become as sin inverse of value, and obtain the result as below,

                       A=41.81^{\circ}

3 0
4 years ago
A
Arlecino [84]

i think the data is not complete but that's according to me

7 0
3 years ago
A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod. relat
Crank
<span>d.rotating counterclockwise and slowing down This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.</span>
7 0
4 years ago
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