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fiasKO [112]
2 years 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]2 years 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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horsena [70]
This would be called the battery. Hope this helps.
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Pls help me prove the decay constant equation
belka [17]
Decay constant, proportionality between the size of a population of radioactive atoms and the rate at which the population decreases because of radioactive decay
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4 years ago
A 1.50 V potential difference is maintained across a 1.50 m wire that has a cross-sectional area of 0.600 mm2. How much power is
Goshia [24]

Given that the potential difference is V = 1.5 V.

The length of the wire is l = 1.5 m.

The cross-sectional area is

\begin{gathered} A\text{ = 0.6 }mm^2 \\ =0.6\times10^{-6}m^2 \end{gathered}

The resistivity of the wire is

\rho\text{ = 5.25}\times10^{-8}\Omega\text{ m }

We have to find the power dissipated in the wire.

First, we need to calculate resistance.

The resistance can be calculated as

\begin{gathered} R\text{ = }\rho\frac{l}{A} \\ =5.25\times10^{-8}\times\frac{1.5}{0.6\times10^{-6}} \\ =0.13125\Omega \end{gathered}

The formula to calculate power is

P\text{ =}\frac{V^2}{R}

Substituting the values, the power will be

\begin{gathered} P\text{ = }\frac{(1.5)^2}{0.13125} \\ =17.1\text{ W} \end{gathered}

Thus, the power dissipated in the wire is 17.1 W

5 0
2 years ago
A 1 cm2 silicon solar cell has a saturation current of 10A and is illuminated with sunlight yielding a short-circuit photocurren
Yuki888 [10]

Answer:

The answer is "13% and 83%"

Explanation:

The highest power is generated for:

\to \frac{dP}{dV_a}=0= I_s(e^{\frac{V_m}{V_t}} -1)-I_{Ph} +\frac{V_m}{V_t} I_s e^{\frac{V_m}{V_t}}

 Where voltage, V_m, was its calculated maximum power wattage.  This same following transcend national stable coins the power output:

\to V_m=V_t \ In \frac{1+\frac{I_{ph}}{I_s}}{1+\frac{V_{m}}{V_t}}

The accompanying consecutive values of V_m are done utilizing iteration and an initial value of 0.5 V:

\to V_m = 0.5,\  0.542,\  0.540 \ V

Efficacy is equivalent to:

\eta = |\frac{V_{m} I_{m}}{p_{in}}| =\frac{0.54 \times 0.024}{0.1}=1.3\%

Currently, I_m was calculated with the formula of voltage V_m (4.6.1) as well as the sun is presumed to produce  100 \frac{mW}{cm^2} of power.

The fulfillment factor is equal to:

fill factor= \frac{V_{m} I_{m}}{V_{oc}I_{sc}}= \frac{0.54 \times 0.024}{0.62 \times 0.025}=83\%

where open circuit tension of equations (4.6.1) and I = 0. is calculated. The current is equal to the total current of the photo.

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3 years ago
PLS HELP :) GIVING BRAINLIEST SIMPLE SCIENCE QUESTION HELPS PLSSSS
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OMG ITS B ITS B I HOPE I HELPED U
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