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4 years ago

Can we use momentum to see how fast the earth is going?

Physics

1 answer:

Kisachek [45]4 years ago

6 0

Yes, if we know the Earth's mass

Explanation:

The momentum of an object is a vector quantity given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

$v=\frac{p}{m}$

Therefore, if we know the momentum of the Earth (p) and we know its mass as well (m), we can solve the formula to find the Earth's velocity.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how hi

sweet-ann [11.9K]

<span>373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km</span>

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3 years ago

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Silver has a mass of 10.5 grams and a volume of 19.3 cm3. What is its density?

Neko [114]

Answer:

<h3>The answer is 0.54 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{10.5}{19.3} \\ = 0.54404145...$

We have the final answer as

Hope this helps you

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3 years ago

Do atoms ever touch

Mumz [18]

Answer:

nope never

Explanation:

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3 years ago

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valentina_108 [34]

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3 years ago

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The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/

velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

To get the true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯, P⃗ =AC¯¯¯¯¯¯¯¯

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector. To get the coordinates of these two vectors, use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

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3 years ago