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>
==========================================
Answer:
The outline of the energy transfer are;
a) Kinetic energy → Clockwork spring → Potential energy
b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Please find attached the drawings of the energy transfer created with MS Visio
Explanation:
The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred
a) The energy transfer diagram for the winding up a clockwork car is given as follows;
Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;
Kinetic energy → Clockwork spring → Potential energy
b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;
Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries
Therefore, we have;
Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
The functions of angles are used to find unknown lengths or angles that can't be measured, in terms of known quantities. The trig functions of angles are ratios of lengths, so they're bare naked numbers without units.
Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²
Answer:
The density of the mixture is 0.55kg/m^3
Explanation:
P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole
PV = nRT
V = nRT/P = 1×8.314×273/100 = 22.70m^3
Mass of N2 = 0.4×28 = 11.2kg
Mass of H2 = 0.6×2 = 1.2kg
Mass of mixture = 11.2 + 1.2 = 12.4kg
Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3
