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3 years ago

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Different units used to describe average velocity can be changed from one to another by the use of

1 answer:

lisov135 [29]3 years ago

4 0

<span>Average velocity can be calculated by determining the total displacement divided by the total time of travel. The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point. Average velocity is different from average speed because it considers the direction of travel and the overall change in position.</span>

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I beg you plz help me asap!!!

Strike441 [17]

Answer:

a.

Explanation:

there would be a new planet is our solar system which could cause different gravitation pull on all the planets also there could be possible be new life form or other valuable metals that haven't been discovered on this planet. hope this helps somewhat

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3 years ago

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Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

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vfiekz [6]

Answer:

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Explanation:

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3 years ago

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The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the

Margaret [11]

Answer:

$v=2019.09\ m.s^{-1}$

Explanation:

Given:

time taken by the sun to complete one revolution, $t=601.2\ hr$

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<u>Therefore, angular speed of rotation of sun:</u>

$\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}$

<u>Now the tangential velocity of the sunspot can be given by:</u>

$v=r.\omega$

$v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}$

$v=2019.09\ m.s^{-1}$

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3 years ago

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Ostrovityanka [42]

Answer:

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