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3 years ago

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Possible reasons why seascorpions went extinct

1 answer:

BabaBlast [244]3 years ago

5 0

Sea scorpions or Eurypterids lived about 251.9 million years ago. They were formidable predators and hunters, but they were wiped out in the Great Permian Mass Extinction, which is also known as the Great Dying (96% of all species on Earth went extinct).The largest species like Jaekelopterus was over 7 feet long! They were mainly thought to go extinct because of a slew of natural disasters that occurred when a comet hit the Earth, as well as increased volcanic activity polluting the seas that Sea scorpions lived in, as well as rising sea temperatures. The Silurian, when these Sea scorpions proliferated was when the water was cooler, holding in more nutrients, allowing both Sea Scorpions and other animals to spread all over the world. But as the oceans became polluted from the volcanoes (and the ash they produced) and the global ocean temperatures began to rise, many animals may not have been able to cope or adapt to the extreme change, becoming helpless in their nutrient deficient water (compared to what they were used to).

I hope this helps!

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3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago