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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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Look at the rock sitting on the hill in the picture above. Gravity should make the rock slide down the hill. What force is actin
nordsb [41]
Look at the rock sitting on the hill in the picture above. Gravity should make the rock slide down the hill. What force is acting to balance gravity,keeping the rock in place? - D. friction
Centripetal force and momentum have to do with movement. Gravity cannot balance gravity. 

5 0
3 years ago
Read 2 more answers
g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl
Fynjy0 [20]

Answer:

v=1.5m/s

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}. Since v=r\omega (the ball rolls without slipping) and for a solid sphere I=\frac{2mr^2}{5}, we have:

mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}

So our translational speed will be:

v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s

6 0
2 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
Convert the following: 52 miles/hour to meters/second
Komok [63]

Answer:

23.246 metres per second

Explanation:

divide the speed value by 2.237

5 0
2 years ago
I really need help i’m in a really big rush
kompoz [17]


1. The speed is at a constant rate same with the acceleration and velocity

2. When there getting close to the top there speed, acceleration, and velocity all start to decrease

3. The speed and acceleration and velocity all on the way down drastically increase number

hope I helped
3 0
2 years ago
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