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igor_vitrenko [27]
3 years ago
9

How is it possible for two objects to have the same momentum, but different velocities?

Physics
1 answer:
coldgirl [10]3 years ago
8 0

Answer: momentum has the same direction as that of velocity but when 2 bodies with the same linear momentum & different velocities it has different masses because a vector quantity is represented by a cross product of mass and velocity of object .

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Which is an example of a mixture?
Doss [256]

Answer:

An example of a mixture would be salt water

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3 years ago
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___________ is the process by which wind removes surface materials.
Travka [436]
The answer is deflation...have a good day

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3 years ago
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A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
3 years ago
Will plane flat mirror will make a real or virtual image?<br> Real<br> Virtual
nikdorinn [45]
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8 0
3 years ago
What eccentricity value results in a circular orbit?
Oxana [17]

Answer:

Zero

Explanation:

Given the equation of an ellipse:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

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e=\sqrt{1-\frac{b^2}{a^2}}

For a circle, we have

a=b

Therefore the eccentricity of a circle is

e=\sqrt{1-\frac{1^2}{1^2}}=0

7 0
3 years ago
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