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4 years ago

12

Calculate the refractive index of the substance if the speed of light in the medium is 210,000 km/s.

1 answer:

Rainbow [258]4 years ago

6 0

0.7 is the correct answer

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What is it called when sediment is dropped and comes to rest?

rosijanka [135]

It's called deposition.

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3 years ago

About three billion years ago, single-celled organisms called cyanobacteria lived in Earth’s oceans. They thrived on the ocean’s

pickupchik [31]

I think its Oxygen.
ancient cyanobacteria produced Earth's first oxygen-rich atmosphere, which allowed the eventual rise of eukaryotes. T<span>he chloroplasts of eukaryotic algae and plants are derived from cyanobacteria</span>

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3 years ago

Read 2 more answers

The two cyclists travel at the same speed on level ground. They approach a low hill and decide to coast up instead of hard pedal

Sergeeva-Olga [200]

Answer:

<u>Option-(E): </u>Bike B; its wheels have a smaller moment of inertia.

Explanation:

The two cyclists travel at the same speed on level ground, as they approach a low hill and decide to coast up instead of hard pedaling. At the top of the hill, as bike "B" will have a larger speed, as compared to the bike "A".Due, to the effect of having smaller moment of inertia, I=m×r².

Having no friction nor air resistance, and all the wheels roll on the ground without slipping. While, the B have larger speed,v due to the fact that bike "B" will have a larger speed,"v" as compared to the bike "A".

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3 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

What are the output waveforms of the following waves, after passing through a transformer?

Ber [7]

The output waveforms after passing through the transformer actually depend on the type of transformer used. It could either be a step-up transformer (steps voltage up), or a step-down transformer (steps voltage down). Both transformers have an output voltage in a form of a sine wave.

8 0

3 years ago