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frosja888 [35]
3 years ago
9

The following is current scientific evidence supporting the nebular theory on the formation of the solar system. the composition

of the inner and outer planets, current observations of star formation, and the motion of the solar system current observations of star formation and the path of comets current observations of the structure and motion of the solar system and the nebular galaxy the composition of the inner and outer planets and the amount of hydrogen and helium found in space
Physics
2 answers:
Anvisha [2.4K]3 years ago
7 0

Answer;

A: the composition of the inner and outer planets, current observations of star formation, and the motion of the solar system

Explanation;

The types of objects found within the solar system provide significant clues and evidence to support the Nebular Theory.

-First; the types of Planets and their distributions: with the Rocky planets being close to the Sun, and Gas Giants planets being far from the Sun, Dwarf Planets or Plutoids, a class of Dwarf planets, are found far from the Sun.

-Comets, asteroids, and meteorites recovered on Earth also provide a number of clues and evidence of Nebular-type development. And the motions of most solar system objects orbit and rotate in an organized fashion.

andreyandreev [35.5K]3 years ago
6 0
All planets orbit the sun in a plane, all the planets orbit the sun in the same direction, most of the planets rotate in the same direction. I'm not sure when and answer ends or begins on your question so you can choose from some of the answers I gave you.
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Why isn’t a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits
Grace [21]

Answer:

The reason the bird is not electrocuted is due to some facts about circuit:

1. Completeness of circuit- This circuit needs to be

complete in order for current to flow. The bird standing on only one wire has not completed the circuit.

2. A potential difference: Another factor deciding

the direction of flow of current is (electric)

potential. Current always flows from a higher

potential to a lower potential. In other words it

can be said that electrons flow from lower

potential to higher one. (the direction of electric

current is opposite to that of the electrons). So

we need the potential difference for current to

flow. The bird standing on only one wire has no potential difference.

3. Path of least Resistance- Factor that decides

the path a current will flow in case of parallel

paths is the (electric) resistance offered by the

path. Current will always flow in the path that

offers least resistance. The leg of a bird has high resistance.

Explanation:

It has no potential difference as both the legs of bird are touching the same wire at same constant potential. ... If the bird would touch the ground while sitting on the wire or flap its wings and touch another electric wire with a different voltage, then it would get shocked and likely die by electrocution.

while in the other hand, the Bird that touches two wires with it wings at the same time will get electrocuted because it has completed a circuit and the its feathers created a potential difference .

8 0
3 years ago
Read 2 more answers
50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t
r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

KE=\frac{1}{2}mv^2

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

KE=\frac{1}{2}mv^2

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
makkiz [27]

Answer:

They move outwards.

They don't intersect each other at any point.

They show the electric field.

Explanation:

8 0
3 years ago
Read 2 more answers
Which current is produced in homes
Natali [406]
Batteries Produce DC current while homes produce AC current.
Hope this helped xxxx
6 0
2 years ago
Read 2 more answers
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