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Sonbull [250]
2 years ago
10

The diagram below shows different weights on the see-saw. Will the see-saw move?

Physics
1 answer:
oee [108]2 years ago
8 0

Answer:

yes it will

Explanation:

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Why are latitude and longitude included on maps
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to locate places on earth

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How energy and power are different ???
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Energy is what makes change happen and can be transferred form one object to another. ... Power is the rate at which energy is transferred. It is not energy but is often confused with energy. The watt is the most commonly used unit of measure for power.

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Barium, a group 2 element, forms an ionic compound with sulfur, a group 16 element. What is the formula for barium sulfide?
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Answer:BaS

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Ba- barium

S-sulfur

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What is the most important mineral​
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2 years ago
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
2 years ago
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