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3 years ago

A. List at least 5 examples of events or factors that could cause extinction. 

Physics

2 answers:

baherus [9]3 years ago

8 0

Disease
Lack of natural reourses
war
asteroid
Polution.

sveticcg [70]3 years ago

4 0

My answer is :
- Pollution
- Epidemic/Disease
- Climate change or global warming
- Habitat Degredation
- Catastrophic event as a volcanic eruption
- Asteroids impacts
- Sea level falls
- tsunamis.

Hope this helps !

Photon

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3 years ago

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Which one of the following is not equivalent to 2.50 miles?

lisabon 2012 [21]

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)$

$x = 13200\,ft$

$x = 1.320\times 10^{4}\,ft$ (Choice A)

mi - in

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)$

$x = 158400\,in$

$x = 1.584\times 10^{5}$ (Choice B)

mi - km

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)$

$x = 4.025\,km$ (Different from Choice C)

mi - yd

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft} \right)$

$x = 4400\,yd$

$x = 4.40\times 10^{3}\,yd$ (Choice D)

mi - cm

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)$

$x = 402500\,cm$

$x = 4.025\times 10^{5}\,cm$ (Choice E)

Choice C is not equivalent to 2.50 miles.

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The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
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$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

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3 years ago

In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in (inclu

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Answer:

Explanation:

When the number of slits increases, the intensity of fringes increases.

So, the fringes appear to be more bright.

As we know that the fringe width is inversely proportional to the number of slits, so as the number of slits increases, the fringe width decreases, hence the fringes are narrower, bright and close together.

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