Kinetic energy because of the wood plank is just still and not moving it’s potential but since it’s asking which one it doesn’t have it doesn’t have kinetic energy cause it’s not moving
Answer:
Average speed = 3.63 m/s
Explanation:
The average speed during any time interval is equal to the total distance travelled divided by the total time.
That is,
Average speed = distance/ time
Let d represent the distance between A and B.
Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,
5.15 = d/t1.
Make d the subject of formula
d = 5.15t1
Let t2 represent the longer time for the return trip at 2.80 m/s . That is,
2.80 = d/t2.
Then the times are t1 = d/5.15 5 and
t2 = d/2.80.
The average speed vavg is given by the following equation.
avg speed = Total distance/Total time
Avg speed = d + d/t1 + t2
Where
Total distance = 2d
Total time = t1 + t2
Total time = d/5.15 + d/2.80
Total time = (2.8d + 5.15d)/14.42
Total time = 7.95d/14.42
Total time = 0.55d
Substitute total distance and time into the formula above.
Avg speed = 2d / 0.55d
Avg Speed = 3.63 m/s
Answer:
Visible Light
wavelength = 4000 - 7000 Angstroms = 400 - 700 milli-microns
1 A unit = 10^-10 m
1 mμ = 10^-9 m
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>
==========================================
Transform boundary
is the answer from stemscopes