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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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I think it is D

Explanation:

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Answer:

Five kingdoms & Seven levels

Explanation:

Robert H. Whittaker proposed five kingdom classical system for living organisms in 1969. These were: Kingdom Monera, Kingdom Protista, Kingdom Fungi, Kingdom Animalia, and Kingdom Plantae. His work was the improvement of earlier work conducted by Carl Linnaeus who proposed three kingdom classification system (Animalia, Plantae, and Protista).

Whittaker's system further categorized each kingdom to seven levels. These were: Kingdom → Phylum  →  Class  →   Order  →   Family →   Genus  →   Species. This categorization was based on similarities in organisms at different levels.

PS: The scientific name of each organism is the combination of genus and family name.

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4 years ago
A car has a distance between axles of 6.2 m and a center of mass located 2.2 m from the front axle. What is the
NARA [144]

Answer:

The correct option is  C

Explanation:

From the question we are told that

    The distance between axles of the car  is  d_a  =  6.2

     The position of the car's center of mass is  D  = 2.2 \ m  \ in \ front \  of  \  the \  axel

Now  we can evaluate the distance of the center of mass to the rear axle as follows

        k  =  d_a  -  D

substituting values

      k  =  6.2 -  2.2

      k  = 4 \ m

assuming the car is   at equilibrium,  taking moment about the center of mass

          W_{front \  axle} *  D  -  W_{rear \  axle} *  k  = 0

=>        W_{front \  axle} *  D     =  W_{rear \  axle} *  k

=>       \frac{W_{front \  axle}}{W_{rear \  axle} }       =    \frac{k}{D}

substituting values

             \frac{W_{front \  axle}}{W_{rear \  axle} }       =    \frac{4}{2.2}

           \frac{W_{front \  axle}}{W_{rear \  axle} }       =    1.81       Note [W_{front \  axle} is the front axle weight  and  

                                                      W_{rear \  axle} is the rear axle weight  ]

6 0
3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

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3 years ago
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