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3 years ago

Which of the following statements is true?

Physics

1 answer:

Grace [21]3 years ago

3 0

I would recommend C)Organic compounds always contain carbon and hydrogen.

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A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an x axis. He can legally

Katyanochek1 [597]

Answer:125.68°

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3 years ago

Dynamo converts the

Sav [38]

Answer:

$\gamma \: dynamo \: converts \: the \: mechanical \: energy \: into \: electrical \: energy \: .$

Explanation:

Hope it helps you :)))

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3 years ago

Read 2 more answers

Select the correct answer.

Aloiza [94]

Answer:

Explanation:

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3 years ago

Velocity is:

sleet_krkn [62]

Answer:

Explanation:

Solution:-

- The Quantity of theory of money states:

M * V = P * Y

Where,

M = Money supply

V = Velocity of money exchange

P = The price level

Y = Real GDP

- By re-arranging the formula and solving for "V" we have:

V = P*Y / M

- The expression on right hand side increases if exchange of dollars increases.

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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