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Nikitich [7]
2 years ago
10

What is the angular momentum at a radius of 2 m with an object of 5 kg at a velocity of 20 m/s?

Physics
2 answers:
aev [14]2 years ago
5 0

The angular momentum is 200 kg m^2 s^-1

<h3>what is angular momentum?</h3>

Angular momentum is the product of  linear momentum and the perpendicular distance. Linear momentum is the product of mass and velocity, where radius is the perpendicular distance

Angular momentum = mass * velocity * radius

Angular momentum = 5 * 2 * 20

Angular momentum = 200 kg m^2 s^-1

read more about angular momentum here: brainly.com/question/4126751

ahrayia [7]2 years ago
3 0

Answer:

200

Explanation:

angular momentum=mvr

=2×5×20

= 200kgm2/s

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The frequency of the light source is 1.5 x 10¹⁵ Hz.

<h3>Frequency of the light source</h3>

The frequency of the light source is determined using the following equations;

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f = (3 x 10⁸) / (2 x 10⁻⁷)

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Learn more about frequency of light here: brainly.com/question/10728818

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A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J. What is the mass of the soccer ball
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A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
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Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

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we wil get the area due to rotation of radius of baton is

\Delta A = \frac{1}{2} \Delta \theta R^2

The  formula for the induced emf is

E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

=B\frac{\Delta  A}{\Delta  t}

B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

so, the induce EMF will be

E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

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