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3 years ago

What are the factors of evaporation and give explanation

2 answers:

7 0

Temperature and rate of evaporation are proportional to each other. Surface area: As the surface area increases, the rate of evaporation increases. The surface area and rate of evaporation are proportional to each other. Humidity: The rate of evaporation decreases with an increase in humidity.

damaskus [11]3 years ago

5 0

Answer:

The three primaryfactors of evaporation are heat, atmospheric pressure (which determines the percentage of moisture) and the movementofair. At the molecular level, there is no strict boundary between the liquid state and the vapor state

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A thin rod rotates at a constant angular speed. Consider the tangential speed of each point on the rod for the case when the axi

ASHA 777 [7]

Answer:

v = R w

With this expression we see that for each point at different radius the tangential velocity is different

Explanation:

They indicate that the angular velocity is constant, that is

w = dθ / dt

Where θ is the radius swept angle and t the time taken.

The tangential velocity is linear or

v = dx / dt

Where x is the distance traveled in time (t)

In the definition of radians

θ = s / R

Where s is the arc traveled and R the radius vector from the pivot point, if the angle is small the arc (s) and the length (x) are almost equal

θ = x / R

We substitute in the speed equation

v = d (θ R) / dt

The radius is a constant for each point

v = R dθ / dt

v = R w

With this expression we see that for each point at different radius the tangential velocity is different

6 0

3 years ago

A mango fruit drops from the top of it's tree which is 40 m. How long does it takes to reach the ground

IrinaVladis [17]

T=distance over speed
T=40m over 9.8ms
T=answer

4 0

3 years ago

True or false? To develop good alternatives, one should brainstorm ideas and consider different perspectives

Contact [7]

From my experience, I would say it is true.

7 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

Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa

Alex

Answer:

$B=1.1636*10^{-3}T$