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3 years ago

When determining the motion of the planets in the solar system, what is a good reference point to use? Explain

2 answers:

7 0

The center of the sun is a good reference point when discussing
the motions of the planets.

When we examine the motions of the planets relative to the sun,
the nature of their elliptical orbits is clearly revealed. We also
see the influence of the planets on other smaller bodies, like
their moons, and the comets and asteroids.

We know that the sun is orbiting the center of the Milky Way Galaxy,
and dragging the whole solar system along with it. But if we ignore
that, and study the motions of solar-system objects relative to the
sun, then the motions are a lot simpler, and easier to understand.

FrozenT [24]3 years ago

3 0

The sun. The planets actually move around the sun in a circumduction system, which is when their orbiting position around the sun while moving through space. The planets are continously moving at a rate concurrent with their proximity in regards to the sun as the sun's gravity pulls them while moving.

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If the elephant were then allowed to fall straight down, how fast would it be moving when it landed back on the ground?

Oxana [17]

Answer:

0 mph because it is probably dead

6 0

3 years ago

The weight of the atmosphere above 1 m- of

mars1129 [50]

Answer:

1.09 kg.m

Explanation:

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2 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago

A theory is a hyothesis that has been varified by multiple investigations. <br> True or false

irakobra [83]

<span>A theory is a hyothesis that has been varified by multiple investigations.

true</span>

7 0

3 years ago

While playing her guitar , karen plucks one string with increasing levels of force. What effect does this have on the sound prod

Lena [83]

Answer:

The amplitude of vibration of string will increase due to which loudness of sound will increase

Explanation:

As we know that the guitar is based on the principle of Resonance. When string of the guitar vibrates at a given frequency then the sound produced in the hollow part of the guitar will also be at same frequency.

This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.

Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.

So here we can say that amplitude and intensity of sound related as

$I = kA^2$

so on increasing amplitude the intensity will increase and hence it will produce loud sound

8 0

3 years ago

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