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Fantom [35]
3 years ago
8

What occurs when vibrations traveling through an object match the objects natural frequency

Physics
2 answers:
Paraphin [41]3 years ago
7 0
It would be Resonance
erik [133]3 years ago
5 0
<span>the answer is - resonance

</span>
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Can a particle with constant speed be accelarating?<br>what if it has constant velocity?​
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Explanation:

it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing.

When an object is moving with constant velocity, it does not change direction nor speed and therefore is represented as a straight line when graphed as distance over time.

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What exercise became part of movement in the 1970
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Aerobic exercise became part of this movement in the 1970s.

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Light travels 3,00,000 km/s . Is it velocity or speed?​
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speed

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according to your question, your answer is speed

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Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:
FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle (v), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

v = R\cdot \omega (1)

Where:

R - Radius, measured in kilometers.

\omega - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that \omega \approx 5235.988\,\frac{rad}{s} and R = 1\times 10^{-4}\,km, then the linear velocity is:

v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)

v = 0.524\,\frac{km}{s}

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

\omega = \frac{2\pi}{T} (2)

Where T is the period, measured in seconds.

If we know that T = 31536000\,s, then the angular speed of the Earth is:

\omega = \frac{2\pi}{31536000\,s}

\omega = 1.992\times 10^{-7}\,\frac{rad}{s}

Now, we determine the linear speed:

v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)

v = 29.88\,\frac{km}{s}

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

6 0
3 years ago
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