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almond37 [142]
2 years ago
9

(MC) What is the result of the Sun's light being refracted when it is high in the sky around noon?

Physics
2 answers:
Ivan2 years ago
5 0

The result of the Sun's light being refracted when it is high in the sky around noon is blue, indigo and violet are strongly bent that mostly skim through the upper atmosphere making the sky appear blue. The refraction of the light during noon makes the sky blue.

Fittoniya [83]2 years ago
5 0

Explanation:

Refraction is the phenomenon which occurs when there is change in the direction of the light while traveling from one medium to the another medium.

Our atmosphere consists of the rarer medium and denser medium. The light gets refracted from when it travels from the sun. Blue, indigo and violet are strongly bent than any other colors. The blue has a shorter wavelength. Blue light is scattered more by the tiny air molecules than any other colors.  Therefore, the sky appear blue in noon.

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Answer:

12 meters per second (12 m/s)

why?

Because if you divide 10 seconds by 10 and 120 by 10, you will get 12 meters in 1 second.

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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

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0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

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3 years ago
A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe
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Answer:

 ω₂=1.20

Explanation:

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mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

I_1=\dfrac{M+m}{2}r^2

I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2

I_1=0.75\ kg.m^2

Final mass moment of inertia

I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

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7 0
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Answer:

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Answer:

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