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luda_lava [24]
4 years ago
7

Why does green light has a lower frequency than blue color?

Physics
1 answer:
Naya [18.7K]4 years ago
5 0
The best way to look at it is the other way around.

Electromagnetic waves exist over a huge range of frequencies, from less
than 1 per second up to more than 3 billion billion billion per second.

If electromagnetic waves with frequencies between maybe about
4 million billion per second up to about 7.5 million billion per second
go into your eye, you SEE them.

Outside of that range, your eye doesn't know that there's anything there.  But
IN that range, your eye has incredible nerve ends in the back that can tell the
difference between different frequencies.  The highest frequencies in that range
produce the feeling of blue in your brain.  The next highest ones make a feeling
of green.  The next highest ones make the impression of yellow, and so on and
so on, down to the lowest frequencies that your eye can detect, and those look
red. If the frequency is any lower than red, you don't see it even when it pours
into your eyes.
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A 0.450-kg hockey puck, moving east with a speed of 5.80 m/s, has a head-on collision with a 0.900-kg puck initially at rest. As
Elodia [21]

Answer:

Explanation:

mass of first puck, m1 = 0.450 kg

initial velocity of first puck, u1 = 5.80 m/s east

mass of second puck, m2 = 0.9 kg

initial velocity of second puck, u2 = 0

Let v1 and v2 are the final velocities of the pucks after collision.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.45 x 5.8 + 0.9 x 0 = 0.45 x v1 + 0.9 x v2

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e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}

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3 years ago
Question No 1 Find the voltage drop across 24 ohm resistor and current flowing through 22 ohm resistor in the given circuit as s
lord [1]

Answer:

8.25 V

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We can ignore the 22Ω and 122Ω resistors at the bottom.  Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.

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This resistance is in parallel with the 24Ω resistor.  The equivalent resistance is:

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8.807Ω + 4Ω = 12.807Ω

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12 V / 12.807Ω = 0.937 A

The voltage drop across the 4Ω resistor is:

(0.937 A) (4Ω) = 3.75 V

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