suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in surrounding air"

Answers

Answer 1

Answer: 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.

Experiment A:

-- 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.
=============================
Show that the 1 ounce of water evaporated faster
when it had more surface area.
============================================
============================================

Experiment B:

-- 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 first section 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.
==========================================
Show that it took longer to evaporate when the air
blowing over it was already loaded with vapor.
==========================================

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An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.5 x 10⁻⁷ C/m², and the plates are separated by a distance of 1.7 x 10⁻² m. How fast is the electron moving just before it reaches the positive plate?

Answers

The speed of the electron before reaching the positive plate is $1.30 * 10^(7)\ m / s$

Explanation:

As per Gauss law of electro statistics, the electric field generated by a capacitor is directly proportional to the surface charge density of the plate and inversely proportional to the dielectric constant. In simple words, the electric field is proportional to the surface charge density. So,

$\text {Electric field}=(\sigma)/(\varepsilon_(0))$

And then from the second law of motion, $F=m * acceleration$

So acceleration exerted by the electrons will be directly proportional to the force exerted on them and inversely proportional to the mass of the electron.

$Acceleration =(F)/(m)$

Since force is also calculated as product of charge with electric field in electrostatic force,

$\text {Acceleration}=(q E)/(m)=(q \sigma)/(m \varepsilon_(0))$

So, the charge of electron $q=1.6 * 10^(-19)\ \mathrm{C}, \sigma=\text { Charge per unit area }=2.5 * 10^(-7)\ \mathrm{C} / \mathrm{m}^(2)$

m is the mass of electron which is equal to $9.11 * 10^(-31)\ \mathrm{kg}$

$\varepsilon_(0)=8.85 * 10^(-12)\ \mathrm{Nm}^(2) \mathrm{C}^(-2)$

Then,

$\text { Acceleration }=(1.6 * 2.5 * 10^(-19) * 10^(-7))/(9.11 * 8.85 * 10^(-31) * 10^(-12))=(4 * 10^(-19-7))/(80.62 * 10^(-31-12))$

$\text { Acceleration }=0.0496 * 10^(-19-7+31+12)=0.0496 * 10^(17)\ \mathrm{m} / \mathrm{s}^(2)$

So the acceleration of the electron in the capacitor will be $4.96 * 10^(15) m / s^(2)$

Then, the velocity can be observed from the third equation of motion.

$v^(2)=u^(2)+2 a s$

As u = 0 and s is the distance of separation between two plates.

$\begin{array}{c}v^(2)=0+\left(2 * 4.96 * 10^(15) * 1.7 * 10^(-2)\right) \nv^(2)=16.864 * 10^(15-2)=16.864 * 10^(13)=1.684 * 10^(14)\end{array}$

Thus, $v=\sqrt{\left(1.68 * 10^(14)\right)}=1.30 * 10^(7)\ m/s$

So, the speed of the electron before reaching the positive plate is $1.30 * 10^(7) \mathrm{m} / \mathrm{s}$ .

The strength of the electric field 0.5 m from a 6 µC charge is N/C. (Use k = 8.99 × 109 N•

Answers

The formula for Electric Field Strength:

E = Q /(4πε₀r²)

Note:

1 /(4πε₀) = k = 8.99 * 10⁹ Nm²/C²

Q = 6 μC = 6 * 10⁻⁶ C. r = 0.5m

E = kQ / r²

E = (8.99 * 10⁹ * 6 * 10⁻⁶) /(0.5²) Use a calculator

E = 215.6 * 10³ N/C

Electric field strength = 215 600 N/C.

I hope this helps.

A mountain climber ascends a mountain to its peak. The peak is 12,470 ft above sea level. The climber then descends 80 ft to meet a fellow climber. Find the climber's elevation above sea level after meeting the other climber. A. -12,390 ft. B. 12,550 ft. C. 11,670 ft. D. 12,930 ft

Answers

The answer is 12,390 ft.

At first, a climber is at 12,470 ft above sea level. But then, he goes down 80 ft to meet a fellow climber. So, this simply needs to be distracted:
12,470 ft - 80 ft = 12,390 ft
This is the elevation above sea level at which he meet the other climber.

This is heat or electrical transfer by contact.

Answers

Answer:

The correct answer for this problem is conduction.

Explanation:

I just did the test and conduction was the correct answe.

If you were to throw flour on the man in the photo, the flour would stick to the man's chest everywhere but on his scar. Why?

Answers

Explanation:

Flour typically sticks to surfaces due to adhesive forces, and it's more likely to stick to areas with moisture or oils on the skin. If the scar on the man's chest lacks moisture or oils (which is common for scars), the flour may have difficulty adhering to that specific area. However, it could stick to the surrounding skin where there might be more moisture or natural oils.

A car is moving in uniform circular motion. If the car's speed were to double to keep the car moving with the same radius, the acceleration wouldincrease by a factor of 2
increase by a factor of 4
decrease by a factor of 2
decrease by a factor of 4