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andriy [413]
3 years ago
13

What happens when visible light strikes a surface?

Physics
1 answer:
Amiraneli [1.4K]3 years ago
6 0
Absorption: 

<span>If the wavelength of the light happens to match the distance between two electron levels in the material, it can be absorbed. Light disappears and its energy becomes part of the heat energy of the material. </span>

<span>Metals have so many free electrons that all wavelengths are absorbed -- they are opaque to xrays, visible light, radio waves, everything. </span>
<span>Pigments have specific electron gaps that absorb only the light of a specific wavelength, which we see as color </span>
<span>Transparent materials do not happen to have any gaps between electron levels that match any of the wavelengths of visible light. </span>
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Calculate the frequency of the wave shown below.
jenyasd209 [6]

\color{skyblue}{ \underline{  \frak { \: option \: ( \: c \: )  =  2 \: hertz   ✓}}}

\:  \:

Given :

  • Wavelength ( λ ) = 2 m

\:  \:

  • Speed = 4 m/s

\:  \:

We, have to find frequency :

\:

  • \large \tt \: Frequency =  \frac{Speed}{Wavelength ( \: λ \: )}

\:  \:

  • \large \tt \: Frequency =  \frac{4}{2}

\:  \:

  • \large \tt \: Frequency = \cancel  \frac{4}{2}

\:  \:

  • \pink{ \boxed{\large \tt \: Frequency =2 \: Hertz ✓}}

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Hope Helps!

5 0
2 years ago
What is the voltage in a circuit if the current is 6.2 A and the resistance is 18 ohms?
Nitella [24]
We already know the formula:
Voltage = Current * Resistance
In the given question, there are numerous information's that are already given.
Current = 6.2 A
Resistance = 18 ohms
Then
Voltage = 6.2 * 18 Volts
             = 111.6 Volt
So, the voltage in the circuit will be 111.6 volts. I hope it helps you.
4 0
3 years ago
Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:
Maru [420]
The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

So, the formula for the instantaneous acceleration is:

a = lim Δv
Δt→0 Δt
8 0
2 years ago
louis Pasteur developed the germ theory of disease, which state that many diseases are caused by microorganismas. Had scientists
Olegator [25]
Sciencefndkdkdkckkfkkdfk
3 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 &lt; r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
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