Answer:
Atoms with the same number of protons but with different electrical charges are different ions
Explanation:
Ions are defined as those atoms or molecules which carry charge
That's two different things it depends on:
-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.
Here's what I have in mind for an experiment to show those two dependencies:
-- a closed box with a wall down the middle, separating it into two closed sections;
-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.
-- a tiny fan that blows air through a tube into the hole in one outer wall.
<u>Experiment A:</u>
-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================
<u>Experiment B:</u>
-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.
A) Conservation of Energy,
![KE = PE](https://tex.z-dn.net/?f=KE%20%3D%20PE)
![\frac{1}{2} mv ^2 = \frac{1}{2} k A^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%20%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20A%5E2)
Here,
m = Mass
v = Velocity
k = Spring constant
A = Amplitude
Rearranging to find the Amplitude we have,
![A = \sqrt{\frac{mv^2}{k}}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cfrac%7Bmv%5E2%7D%7Bk%7D%7D)
Replacing,
![A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.750%29%2831%2A10%5E%7B-2%7D%29%5E2%7D%7B13%7D%7D)
![A = 0.0744m](https://tex.z-dn.net/?f=A%20%3D%200.0744m)
(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.
The Period is defined as
![T = 2\pi \sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
Replacing,
![T = 2\pi \sqrt{\frac{0.750}{13}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7B0.750%7D%7B13%7D%7D)
![T= 1.509s](https://tex.z-dn.net/?f=T%3D%201.509s)
Now the velocity is described as,
![v = \frac{2\pi}{T} * \sqrt{A^2-x^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%7D%7BT%7D%20%2A%20%5Csqrt%7BA%5E2-x%5E2%7D)
![v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%7D%7BT%7D%20%2A%20%5Csqrt%7BA%5E2-0.75A%5E2%7D)
We have all the values, then replacing,
![v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B1.509%7D%5Csqrt%7B%280.0744%29%5E2-%280.750%280.0744%29%29%5E2%7D)
![v = 0.2049m/s](https://tex.z-dn.net/?f=v%20%3D%200.2049m%2Fs)
elasticity stretches and can also return to it's normal size ..
Answer:
v = 2,99913 10⁸ m / s
Explanation:
The velocity of propagation of a wave is
v = λ f
in the case of an electromagnetic wave in a vacuum the speed that speed of light
v = c
When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium
λ = λ₀ / n
where n is the index of refraction of the material medium.
Therefore the expression is
v =
Let's look for the frequency of blue light in a vacuum
f =
f =
f = 6.667 10¹⁴ Hz
the refractive index of air is tabulated
n = 1,00029
let's calculate
v =
450 10-9 / 1,00029 6,667 1014
v = 2,99913 10⁸ m / s
we can see that the decrease in speed is very small