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olya-2409 [2.1K]
3 years ago
8

A peach has a layer of skin, a thick section of fruit, and a pit in the center. Which of these would a peach be a good model for

?
A.
the ocean
B.
the Sun
C.
the solar system
D.
the Earth
Physics
2 answers:
timofeeve [1]3 years ago
8 0

Answer:

The earth

Explanation:

Vikentia [17]3 years ago
3 0

Answer:

D. the Earth

Explanation:

The Earth has a solid metal core at the center, which is similar to the pit of a peach. The Earth also has many thick layers of rock and liquid metal between the surface and the center of the Earth, similar to how a peach has a layer of skin and a thick section of fruit. The other answer choices don't have any similarities to a peach. The ocean is entirely liquid, the sun is a ball of gas, and the solar system contains lots of different planets and stars that all are made up of very different things.

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</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon

If there are given values, substitute in the equation and solve for the torque.</span>
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arlik [135]

thats how it works and thanks for points

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love history [14]

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the answer is D

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If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :
gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

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So

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5 0
2 years ago
Read 2 more answers
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
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