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3 years ago

Astronomers estimate that a galaxy contains a. 100 stars. c. millions or billions of stars. b. 1 star. d. 100,000 stars.

Physics

1 answer:

In-s [12.5K]3 years ago

3 0

C.

Galaxies are very big, with millions or billions of stars in them.

Hope this helps!

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If the speed of light in air is 3.00 times 10 to the 8 m/s power and the speed of light in water is 2.25 times 10 to the 8 power

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I also attached some of my physics notes on refraction and that sort of jazz as they may help :)

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3 years ago

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3 years ago

A child wants to pump up a bicycle tire so that its pressure is 1.2 × 105 pa above that of atmospheric pressure. if the child us

WINSTONCH [101]

General expression is;
Pressure = Force/Area
In which,
Pressure = Required pressure + Atmospheric pressure = (1.2*10^5) + (101325) = 221325 Pa = 221325 N/m^2

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3 years ago

Read 2 more answers

a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?

evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

$\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}$

So,

$h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}$

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0

4 years ago