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2 years ago

From what three regions do most of the small objects in the universe come?

Physics

1 answer:

Ivenika [448]2 years ago

3 0

Answer:

Scientists who study the Sun usually divide it up into three main regions: the Sun's interior, the solar atmosphere, and the visible "surface" of the Sun which lies between the interior and the atmosphere. There are three main parts to the Sun's interior: the core, the radiative zone, and the convective zone.

Explanation:

Hopefully this helps :)

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Paleontologists are experts on the rock cycle true or false

kow [346]

Answer:

I cannot do the laws of physics cuz I hate science but I'm just dance

Explanation:

it's it's true

8 0

3 years ago

Why is gravitational forces equal for everyone?

Alik [6]

That equation is Newton's universal law of gravitation. ... Any two masses exert equal-and-opposite gravitational forces on each other. If we drop a ball, the Earth exerts a gravitational force on the ball, but the ball exerts a gravitational force of the same magnitude (and in the opposite direction) on the Earth.

7 0

3 years ago

A 40-w lightbulb connected to a 120-v source experiences a voltage surge that produces 132 v for a moment. by what percentage do

andrey2020 [161]

R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%

3 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to Newton's 2nd law of motion:

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

3 years ago

The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

astra-53 [7]

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

4 0

2 years ago