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Kruka [31]
4 years ago
9

Why are large astronomical bodies round

Physics
2 answers:
anastassius [24]4 years ago
8 0

I would have to go with gravity, as each object in this universe has some amount of gravity, even humans and creatures and plants.

when a rock is thrown into space, it will have to travels light years away and pull in thousands of other asteroids into yet sometimes this doesn't work and sometimes nothing will happen depending on the speed of how they are going. 

collision of these rocks are also another cause to the round shape, remember these rocks could be thousands of millions of years old. and this would be easy for them to shape because of the amount of asteroids in space.

this shape is also by the gravity on it in of from it. though this rare and it will be called a meteorite.

-Dominant- [34]4 years ago
7 0
<span>Matter of all types have gravity, which causes it to attract to each other. The most efficient way for all this matter to congregate is the sphere. As they consolidate, they form the shape. As they compress, temperatures in the center of the mass start to go up and if it hits the proper point, it can ignite and become a star.</span>
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According to the Theory of Plate Tectonics, which of the following statements is correct?
kramer

Answer:

It's D. The geothermal energy from Earth's core loses most of its heat before reaching the outer layer. The plates on the outer layer have been proven to be moving due to evidence such as earthquakes, volcanic eruptions, and tsunamis.

Explanation:

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3 years ago
Question 5 of 10
jonny [76]

Answer:

A. The particle model, because only high-energy frequencies of light  can remove electrons .

Explanation:

Each photon of blue light has higher energy than each photon of red light has  . So when each photon strikes each electron , it gets ejected . But the photon of red light has not sufficient energy to eject electron . Once the photon of red light strikes the electron , the energy is wasted off . Energy of photon can not be accumulated . Thus photon behaves like particle .

4 0
3 years ago
How would I workout question 4?? Please help
Agata [3.3K]

Answer:

550 kg

Explanation:

mass = E / gh

= 33000/60

=550

plzzz......

mark it as a brilliant answer

3 0
3 years ago
Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?
lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

  • F₁ = 225.4 N
  • F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

           

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0\frac{W_1  \ 1 + W_2 \ 1.5}{2}

Let's calculate F₂

         F₂ = \frac{W_1 \ 1 + W_2 \ 1.5 }{2}  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = \frac{(12 + 68 \ 1.5 ) \  9.8}{2}  

         F₂ = 558.6 N

We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

         F₁ = (12 +68) 9.8 - 558.6

         F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

  • F₁ = 225.4 N
  • F2 = 558.6 N

Learn more here:  brainly.com/question/12830892

5 0
2 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

3 0
4 years ago
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