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3 years ago

What is salinity ? ..

Physics

2 answers:

padilas [110]3 years ago

8 0

Answer:

Salinity is the amount of salt in grams present 1000 grams of water. The Average salinity of the oceans is 35 parts per thousand .

<h3>Hope this helps you .</h3>

Sry for short answer .

AveGali [126]3 years ago

7 0

Answer:

<h3>Salinity is the saltiness or amount of salt dissolved in a body of water, called saline water. It is usually measured in g/L or g/kg. </h3>

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Trumpeter A holds a B-flat note on the trumpet for a long time. Person C is running towards the trumpeter at a constant velocity

Vikki [24]

You didn't mention it, but the trumpeter herself has to be standing still.

<span>Person C, the one running towards the trumpeter, hears a pitch
that is higher than B-flat. (A)

Person B, the one running away from the trumpeter, hears a pitch
that is lower than B-flat.

Person D, the one standing still the whole time, hears the B-flat.</span>

5 0

3 years ago

Read 2 more answers

HELPPP IM DESPERATE

damaskus [11]

the first one cuz I know

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3 years ago

2<br><br> How does Descartes' "quality of motion" differ from the modern<br><br> momentum?

Rzqust [24]

Answer: Descartes was more of speed which defers from modern day velocity.

Explanation:

Descartes law if conservation referred or defined “motion” rather than “momentum” as what is obtainable in today's world as ”speed” the rate at which something moves rather than “velocity” which is a product of speed and direction. So in conclusion Descartes was more of speed which defers from modern day velocity.

6 0

3 years ago

What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0

3 years ago

A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$