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fiasKO [112]
1 year ago
9

Sunlight shines through a salt shaker. Should you expect more, less, or the same amount of scattering from blue light compared t

o red light? how about for light shining through a plume of smoke?
Physics
1 answer:
pshichka [43]1 year ago
4 0

Blue light will scatter more compared to red light.

Blue light has a short wavelength; red light a longer wavelength. The sky looks blue because blue light is scattered far more than red light, owing to the shorter wavelength of blue light.

<h3>What is scattering of light?</h3>

Scattering of light is the phenomenon in which light rays deviate from their original path upon striking an obstacle like dust, gas molecules or water vapors. Scattering of light gives rise to many spectacular phenomena such as Tyndall effect and the red hues that can be seen at sunrise and sunset.

<h3>What is the scattering of light with example?</h3>

Some example of scattering of light that we come across in day-to-day life are: Blue colour of the sky: Out of the seven components present in sunlight, blue colour is scattered the most by the particles present in the atmosphere and hence, the sky appears blue.

To learn more about scattering of light visit:

brainly.com/question/9922540

#SPJ4

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The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

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A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi
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Answer:

Equilibrium temperature will be T=52.2684^{\circ}C

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So 2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)

0.334T-3.67=1688-32.031T

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