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stepan [7]
3 years ago
6

What factors have caused Earth to evolve over geologic time?

Physics
1 answer:
kow [346]3 years ago
8 0
Earth, the planet:

1. Collision of Earth with Mars-sized planet.

2. Formation of the Moon and its gravitational effects.

3. Cooling and solidification of Earth's molten outer surface.

4. Period of intense asteroid bombardment.

5. Build-up of liquid water on Earth's surface.

6. Appearance of continents.

7. Start of plate tectonics.

8. Appearance of photosynthetic bacteria.

9. Change in atmosphere due to photosynthetic bacteria.

10. Widespread appearance of complex organisms.

11. Adaptation of plant and animal life to land.

12. Periods of intense, long-lasting volcanic activity.

13. Re-positioning of Earth's continents and oceans, and its

resultant effect on weather patterns and ocean circulation.

14. Changes in Earth's magnetic fields.

15. Changes in Earth's orientation to the sun.

16. Random devastating meteorite impacts.

17. Glaciations, floods, earthquakes.

Life: Abundance, Diversity and Extinction of different species in response to changing climatic, ecologic and edaphic conditions.
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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
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To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

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The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

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\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

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\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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Explanation:

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