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3 years ago

Place these bodies of our solar system in the proper order of formation.

Physics

2 answers:

Westkost [7]3 years ago

5 0

Answer:

solar nebula, the sun, planetesimals, inner planets, and last but not least outer planets

Dahasolnce [82]3 years ago

5 0

Answer:

1. Solar Nebula

2. The Sun

3. Planetesimals

4. Inner planets

5. Outer planets

Explanation:

A star is formed in a molecular cloud of gas and dust, mainly composed of hydrogen and helium. The Nebular Theory establishes, for the formation of the solar system, that the cloud starts to collapse under its own gravity when it receives a shock wave from a near event, for example, a supernova explosion. That results in the cloud breaking in small pieces, and those pieces constitute a possible future star.

Then it begins to accrete and rotate as a consequence of the angular momentum. In the center of that disk when it reaches the necessary temperature and pressure a protostar will born.

Around the star, in this case the Sun, fragments of dust combine until they get a meaningful size (planetesimals). According with chemical distribution on the disk of the future solar system, rocky and iron were closer to the Sun and gasses and ice were in the outer part of disk. That may explain why the inner planets are terrestrial and the outer planets are giant gasses.

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A student notices that an inflated balloon gets larger when it is warmed by a lamp. Which best describes the mass of the balloon

nirvana33 [79]

It gets larger because
well let me give you an example
so today in class we looked at a lava lamp with wax inside and there was a lightbulb at the bottom.
we watched as the wax floated up because the molecules inside the wax spreads out and makes the wax less dense.
the wax floats up because (which is related to the balloon getting bigger) the wax is getting less dense and the particles get bigger which ALSO makes the wax less dense.
hope this helps and hope you can relate it to your problem! say thanks if I did help AT ALL! :)

7 0

4 years ago

Read 2 more answers

An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force

lana66690 [7]

Answer:

The friction force acting on the object is 7.84 N

Explanation:

Given;

mass of object, m = 4 kg

coefficient of kinetic friction, μk = 0.2

The friction force acting on the object is calculated as;

F = μkN

F = μkmg

where;

F is the frictional force

m is the mass of the object

g is the acceleration due to gravity

F = 0.2 x 4 x 9.8

F = 7.84 N

Therefore, the friction force acting on the object is 7.84 N

5 0

4 years ago

Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up

loris [4]

Puto chupame el semen ok? right?

8 0

3 years ago

A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from

Harman [31]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

$F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s$

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

5 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

4 years ago