Answer:
The percentage of an iceberg submerged beneath the surface of the ocean = 89.67%
Explanation:
Let V be the total volume of the iceberg
Let x be the volume of iceberg submerged
According to Archimedes principle,
weight of the iceberg = weight of the water displaced (that is, weight of x volume of water)
Weight of the iceberg = mg= ρ(iceberg) × V × g
Weight of water displaced = ρ(fluid) × x × g
We then have
ρ(iceberg) × V × g = ρ(fluid) × x × g
(x/V) = ρ(iceberg) ÷ ρ(fluid) = 916.3 ÷ 1021.9 = 0.8967 = 89.67%
Hope this Helps!!!!
Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Answer:
Explanation:
Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this: 
where x is the position, t is the time a is the acceleration and 
is initial velocity. In this way acceleration can be found. 
.
Now we are able to found velocity at any time with the formula: 
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>
==========================================
