All categories

3 years ago

The longest wavelength within the visible spectrum is _______ light.

2 answers:

5 0

The longest wavelength within the visible spectrum is the red light. The answer is letter C. It is called visible light because it is the only light that can be seen by the human eye. Red light is the longest wavelength around 620 to 750 nanometer. It is followed by orange which has a wavelength of 590 t 620 nanometer. And then blue which has a wavelength of 450 to 495 nanometer. And the shortest wavelength is violet which has a wavelength of 380 to 459 nanometer.

Alik [6]3 years ago

5 0

Answer:

red light.

Explanation:

You might be interested in

Frequency is the ____________ that move past a point during a specific amount of time. Frequency is measured in ____________ , a

yulyashka [42]

Answer:

Frequency is the number of waves that move past a point during a specific amount of time. Frequency is measured in Hertz, and is classified as high, medium, or low. Frequency is interpreted as the pitch of a sound. Intensity refers to the loudness of a sound and is measured in decibels. Louder sounds increase the rate of nerve signals relayed to the brain.

Explanation:

6 0

4 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers

A box has a length of 18.1 cm and a width of 19.2cm tall ?

ladessa [460]

Volume = l*w*h = (18.1 cm)(19.2 cm)(20.3 cm) = 7,054 cm^3.

3 0

3 years ago

Which objects are opaque? Check all that apply.

Vinvika [58]

textbook
track shoes
basketball

7 0

3 years ago

Read 2 more answers

Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?A. Fluorescent lamps are more

mr_godi [17]

Answer:

B. Fluorescent lamps operate at a higher temperature than incandescent

Explanation:

Fluorescent lamps have a number of advantages over incandescent lamps which are given in the options given in A, C and D. The option available in B is a drawback, not an advantage. This is because it can give out and radiate more heat as a result of working at a higher temperature. Hence B option is correct.

8 0

3 years ago