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>
==========================================
Are there supposed to be multiple choices for this question?
Number of miles that marker shows when passes through town= 160 miles.
Number of miles that marker shows currently to John = 115 miles.
We need to find the distance between town and John's current location.
For the problem, we can clearly see that Town is at 160 miles away but when John passes the marker shows 115 miles.
So, it's just the difference between 160 miles and 115 miles.
In order to find that difference, we need to subtract those two numbers.
160miles - 115miles = 45 miles.
So, we could say the distance between town and John's current location is 45 miles.
Answer:
d=360 miles
Donna lives 360 miles from the mountains.
Explanation:
Conceptual analysis
We apply the formula to calculate uniform moving distance[
d=v*t Formula (1)
d: distance in miles
t: time in hours
v: speed in miles/hour
Development of problem
The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:
travel data to the mountains: t₁= 8 hours , v=v₁
d= v₁*t₁=8*v₁ Equation (1)
data back home : t₂=4hours , v=v₂=v₁+45
d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)
Equation (1)=Equation (2)
8*v₁=4v₁+180
8*v₁-4v₁=180
4v₁=180
v₁=180÷4=45 miles/hour
we replace v₁=45 miles/hour in equation (1)
d=8hour*45miles/hour
d=360 miles
Answer:
Speed; v = 17 m/s
Explanation:
We are given;
Radius; r = 110m
Angle; θ = 15°
Now, we know that in circular motion,
v² = rg•tanθ
Thus,
v = √(rg•tanθ)
Where,
v is velocity
r is radius
g is acceleration due to gravity
θ is the angle
Thus,
v = √(rg•tanθ) = √(110 x 9.8•tan15)
v = √(288.85)
v = 17 m/s
