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m_a_m_a [10]
4 years ago
10

Why are the four inner planets in our solar system dense and rocky while the outer planets are gaseous?

Physics
1 answer:
iren2701 [21]4 years ago
7 0
I believe it is c. Because the planets were formed billions of years ago out of their respective "ingredients"

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If a 12 kg cat is sitting 5 m up in a tree, how much PE does it have?
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Answer:

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3 years ago
A group of students are provided with three objects all of the same mass and radius. The objects include a solid cylinder, a thi
SOVA2 [1]

Answer:

Sphere, cylinder    hoop

Explanation:

To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy

In the highest part of the ramp

     Em₀ = U = mg y

In the lowest part

Here the energy has part of translation and part of rotation

      E_{mf}  = K_{T} + K_{R}

      E_{mf}  = ½ m v_{cm}² + ½ I w²

Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass

     v_{cm} = w r

    w = v_{cm} / r

Let's replace

   E_{mf} = ½ I (v_{cm} / r)²

Energy is conserved

   mg y = ½ m v_{cm}² + ½ I v_{cm}² / r2

   ½ (m + I / r²) v_{cm}² = m g y

   ½ (1 + I / m r²) v_{cm}² = g y

   v_{cm} = √ [2gy / (1 + I / mr²)]

This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition

   v = d / t

   t = d / v

   t = d / (√ [2gy / (1 + I / mr²)])

   t = (d / √ 2gy) √(1 + I / m r²)

Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.

The moments of inertia of

Sphere is   Is = 2/5 M r²

Cylinder    Ic = ½ M r²

Hoop         Ih = M r²

Let's replace each one and calculate the time

Sphere

    ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)

    ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)

    ts = (d / √ 2gy)      1.1

Cylinder

    tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)

    tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5

    tc = (d / √ 2gy)    1.2

Hoop

    th = (d / √2gy) √ (1 + mr² / mr²)

    th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2

    th = (d / √ 2gy)  1.41

We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is

         ts < tc < th

     Sphere, cylinder    hoop

5 0
3 years ago
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