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3 years ago

What color is the sky really? A. Blue B.Green C.Red D.Oarnge E. ALL of the above

Physics

1 answer:

Nataly [62]3 years ago

7 0

Answer:

I suppose it comes to the time of day, but technically I would suppose (E: All of the Above)

Explanation:

The color of the sky is due to a phenomena called Rayleigh scattering. In Rayleigh Scattering, light waves travel through a medium (in this case, the atmosphere) and strike the individual particles which bend the light. The light waves from the sun are smaller than the particles in the atmosphere (the molecules and the atoms in our atmosphere) and are bent when they strike these particles, resulting in the color of the sky you see at whatever time. I'm no physics major, but this is one of the major reasons we see the sky in the way we do. If you wanted to go further, you could even go and say that the Martian atmosphere is red because of the exact same phenomena.

Did you Know? A Martian sunset is blue, check it out: https://www.bing.com/images/search?q=martian+sunset&FORM=HDRSC2

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

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3 years ago

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asambeis [7]

Answer:

Explanation:

The problem provides you with thesolubility of potassium chloride, KCl , in water at 20∘C , which is said to be equal to 34 g / 100 g H2O . This means that at 20∘C , a saturated solution of potassium chloride will contain 34 g of dissolved salt for every100 g of water.

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Answer:

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goldfiish [28.3K]

Answer:

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