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3 years ago

Why does the frequency of a wave increase as the wavelength decreases

2 answers:

5 0

Because their PRODUCT is always the same number ... the wave speed. So if one of them changes, the other one has to change in the opposite direction, in order to keep their product the same.

olchik [2.2K]3 years ago

3 0

Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

Hope you understood

You might be interested in

In the Millikan oil drop experiment the measured charge of any single droplet was always a whole number multiple of -1.60 x 10-1

BaLLatris [955]

Answer:

Number of electrons, n = 6

Explanation:

Total charge in a single droplet, $q=-9.6\times 10^{-19}\ C$

The measured charge of any single droplet, $e=-1.6\times 10^{-19}\ C$

Let n is the number of excess electrons are contained within the drop. According to the quantization of charge :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}$

n = 6

So, there are 6 electrons contained within the drop. Hence, this is the required solution.

7 0

3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

Mirrors produce images by doing which of the following to light?

Gemiola [76]

Answer: C

Reflection

Explanation: Light travels in a straight line. Reflection is one of the properties of light. And this is the property in which mirror make use of. The ability of light to bounce back. It's this bouncing back characteristics of light ray that eventually produce the image of an object by the mirror.

If the light ray is absorbed, no image will be produced.

4 0

3 years ago

Is it possible to put a distance versus time graph to be universal go mine

xz_007 [3.2K]

Yes I'm pretty sure you can

3 0

3 years ago

How does the size of a planet affected the amount and type of gas in its atmosphere

dalvyx [7]

There are lots of variables that directly and indirectly contribute to the presence of gas on a surface
if the size of a planet is relatively small it will in turn be that of a smaller area which results in the less area to be covered for gas which basically means higher presence
I can go in depth more but I don't think that would be necessary all you need to know is this ...based on the size and gas will in turn be parallel to it's conformity

3 0

3 years ago