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4 years ago

As you watch the surfers right away towards the shoreline what is the shoreline

Physics

1 answer:

Sophie [7]4 years ago

3 0

The shoreline would be your reference point.

A reference point, in physics, is some point in space used to illustrate the location or position of other things, relative to this reference point. So in this case, the shoreline is considered as the reference point to locate the position of the surfers.

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How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it

Oksi-84 [34.3K]

Answer:

Explanation:

more the density , more will be the buoyant force acting on it , less the density less will be the buoyant force acting on it. This is why people float in dead sea and sink in other seas

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3 years ago

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth

Semmy [17]

Answer:

$\lambda=6.83\times 10^{-5}\ m$

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m$

So, the wavelength of the infrared radiation is $6.83\times 10^{-5}\ m$ .

5 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$