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Amiraneli [1.4K]
2 years ago
7

Someone help me answer this.

Physics
1 answer:
horrorfan [7]2 years ago
8 0

The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

<h3>What is angular velocity?</h3>

The velocity of a particle when moving in the circular path.

Let speed of the bug with respect to ground is u.

Speed of bag with respect to ring will be

v = u - (- Rω) =

Then, u = v- Rω...............(1)

Angular momentum of ring and bug will remain conserved.

Initial  momentum: L ring + Lbug =0

Final  momentum:  -2m₁ R²ω + m₂uR =0...............(2)

Using equation (1) and (2), the angular velocity expression will be

ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction

Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

Learn more about angular velocity.

brainly.com/question/17592191

#SPJ1

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A small insect is placed 3.75 cm from a +4.00 −cm -focal-length lens. Calculate the angular magnification.
aev [14]

Explanation:

It is given that,

Object distance, u = -3.75 cm  (negative always)

Focal length, f = +4 cm

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\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

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Angular magnification is as the ratio of image distance to the object distance i.e.

m=\dfrac{v}{u}

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So, the angular magnification of the insect is 16. Hence, this is the required solution.

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3 years ago
If one mass is much greater than the other, the larger mass stays essentially at rest while the smaller mass moves toward it. Su
sukhopar [10]

Answer:

The speed of orbit when it crosses Mercury, v = 5.4 x 10⁴ m/s

Explanation:

Given,

The mass of the comet, m = 1.5 x 10³³ Kg

The mass of the sun, M = 1.99 x 10³⁰ Kg

The velocity of the comet at the orbit of mars, V = 3.1 x 10⁴ m/s

The radius of orbit of mars, R = 2.28 x 10⁸  km

                                                 = 2.28 x 10¹¹ m

The radius of orbit of mercury, r = 5.79 x 10⁷ Km

                                                      = 5.79 x 10¹⁰ m

The velocity of orbit at a distance R from the sun is given by the formula

                                            V = \sqrt{\frac{GM}{R} }

Substituting the given values in the above equation

                                           V = \sqrt{\frac{6.67X10^{-11}X1.99X10^{30}}{2.28X10^{11}}}

                                                    = 2.41 x 10⁴ m/s

Given that the velocity of a comet passing the orbit of mars is 3.1 x 10⁴ m/s

The discrepancy in the velocity is 0.7 x 10⁴ m/s

This is due to the high eccentricity of the orbit of a comet.

The velocity of a comet crossing the orbit of mercury is

                                             v = \sqrt{\frac{6.67X10^{-11}X1.99X10^{30}}{5.79X10^{10}}}    

                                                     = 4.7 x 10⁴ m/s

Adding the discrepancy to the above value gives

                                                  v =  5.4 x 10⁴ m/s

Hence, the velocity of the comet crossing the orbit of mercury is, v =  5.4 x 10⁴ m/s    

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