The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s
<h3>What are the components of velocity?</h3>
We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.
Hence;
Vertical component = 40 m/s sin 15 degrees = 10.35 m/s
Horizontal component = 40 cos 15 degrees = 38.6 m/s
Learn more about components of velocity:brainly.com/question/14478315
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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>
==========================================
Edit: You do mean Ridge?
Rocks near Mid-Ocean Ridge are younger than rocks near the trenches.
Seismic data shows oceanic crust is sinking into the mantle at the trenches.
Matching bands of magnetic rock are found on either side of the Ridge. Earth's magnetic fields change these bands over time.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
Answer: V1 = 3.559 - 0.744V2
Explanation: Given that the
bumper car A has
Mass M1 = (281 kg) moving with
Velocity U1 = 2.82 m/s
bumper car B has
Mass M2 = (209 kg) moving with
Velocity U2 = 1.72 m/s
Where U1, U2 are the initial velocity of the two cars
Since the collision is elastic, we will use the formula below,
M1U1 + M2U2 = M1V1 + M2V2
Substitute the values into the formula
281×2.28 + 209×1.72 = 281V1 + 209V2
640.68 + 359.48 = 281V1 + 209V2
1000.16 = 281V1 + 209V2
Make V1 the subject of formula
281V1 = 1000.16 - 209V2
V1 = 1000.16/281 - 209V2/281
V1 = 3.559 - 0.744V2
Therefore, the velocity of car A which
is V1 after the collision will be expressed as V1 = 3.559 - 0.744V2
