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Valentin [98]
3 years ago
12

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

rrounding air"
Physics
1 answer:
gulaghasi [49]3 years ago
6 0
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>
==========================================
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How to measure the volume of a baseball bat ( need answers ASAP )
vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water.  But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.  

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly.  Some water will run over the top and out of the tub.  That's OK ... it's exactly what you want.  If NO water runs over the top, pour some more in, until it runs out and then stops.  You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan.  The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string.  Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it.  The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub.  Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub.  Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it.  Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions.  Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces.  But you can find all those conversion factors with a search on Floogle.

8 0
3 years ago
Momentum,
Serggg [28]

Answer:

THE WALL MOVES AWAY FROM THE BALL

Explanation:

NEWTON'S THIRD LAW STATES THAT THERE IS A OPPOSITE REACTION

5 0
3 years ago
The two ends of a coiled wire are connected to the electrodes of a lightbulb.
barxatty [35]

Answer:

a

Explanation:

The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current.

in other words, the magnets motion creates a current in the loop

5 0
3 years ago
Two similar cars are driving on a hill. The red car is traveling faster than the blue car. How could the blue car have more tota
yKpoI14uk [10]
D if the blue car started higher it would have more energy but since the red car is lower it is going faster because it’s going down a hill
8 0
3 years ago
Problem 02.061 For the given circuit, assume vS = 10 V, R1 = 9 Ω, R2 = 4 Ω, R3 = 4 Ω, R4 = 5 Ω, and R5 = 4 Ω. Reference Book &am
stich3 [128]

Answer:

Ws=8.75 Watts

Explanation:

As per fig. of prob 02.061, it is clear that R5 and R4 are in parallel, its equivalent Resistance will be:

\frac{1}{Req45}=\frac{1}{R4}+\frac{1}{R5}

\frac{1}{Req45}=\frac{1}{5}+\frac{1}{4}

\frac{1}{Req45}=0.2+0.25=0.45\\ Req45=2.22

Now, this equivalent Req45 is in series with R3, therefore:

Req345=R3+Req45\\Req345=4+2.22\\Req345=6.22

This Req345 is in parallel with R2, i.e

Req2345=(R2^{-1}+Req345^{-1}  )^{-1}\\ Req2345=(4^{-1}+6.22^{-1}  )^{-1} \\Req2345=2.43

Now this gets in series with R1:

Req12345=R1+Req2345\\Req12345=9+2.43\\Req12345=11.43

Now, the power delivered Ws is:

Ws=Vs*I=\frac{Vs^{2}}{Req}  \\Ws=\frac{10^{2} }{11.43} \\Ws=8.75 Watts

8 0
3 years ago
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