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vovangra [49]
4 years ago
14

6. How does the shape and size of the continents today help support the Theory of Continental Drift?

Physics
2 answers:
Ahat [919]4 years ago
5 0
All the continents were connected in one large landmass.If you take a map and cut out all the Continent, you can see that they fit together almost perfectly, like a giant puzzle. This idea supports the fact that <span>all the continents were connected in one large landmass.
</span>hope it helped :)

Lubov Fominskaja [6]4 years ago
4 0
The t<span>heory of Continental Drift is a theory which earth used to have one giant landmass called Pangea. 

The shape and size of today's continents all look like they fit together like a puzzle. The edges look somewhat like they would fit together. If you cut out all continents and pieced them together into one giant landmass it would look nearly perfect. This is why today's continents support the t</span><span>heory of Continental Drift

Hope this helped. Have a great day!</span>
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Lawson's criterion states that the product of ion density and confinement time must exceed a certain number before a break-even
LekaFEV [45]

According to Lawson's criterion, the outcome is determined by the product of ion density and confinement time because the temperature must be maintained for a sufficient confinement time and with a sufficient ion thickness to obtain a net gain of power from a fusion reaction.

<h3>What are Lawson's criterion?</h3>
  • The overall conditions that must be met in order to produce more energy than is required for plasma heating are usually expressed in terms of the product of ion density and confinement time, a condition known as Lawson's criterion.
  • In nuclear fusion devices, confinement time is defined as the amount of time the plasma is kept at a temperature above the critical ignition temperature.
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7 0
1 year ago
A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
jeka94

Answer:

a)  t_l - t_r = 12.54 us

b)  (t_l - t_r) / T = 0.0157  

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

                                      d_l = sqrt ( 2.6^2 + 0.15^2)

                                      d_l = sqrt ( 6.7825 )

                                      d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

                                      t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

                                      t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)  

                                      t_l - t_r = ( 0.0043 / 343 )

                                      t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

                                      T = 1 / f

                                      T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

                                      (t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

                                      (t_l - t_r) / T = 0.0157  

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