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In this video, we're going to talk about PoCA's alarm and how it relates to pressure and fluid flow.
Now, let's consider a pipe that contains a fluid with no viscosity.
If there's no viscosity, then there's no internal friction.
And as a result, the fluid can move with constant speed without any net force, so the net | |
e is zero.
Now, pressure and force are related.
So, if the net force is zero, then the pressure difference has to be zero.
Pressure is forced divided by area.
So, the pressure at point A, if it's 100,000 pass gals,
and if the pressure at point B is the same, then the fluid will continue to flow at constant speed,
if | |
s no friction.
Now, in real life situations, we know this sum level of internal friction.
And so, in order for the fluid to travel at constant speed,
there must be some sort of pressure difference.
And that situation is described by PoCA's alarm.
So, PoCA's law states that the volume flow rate is equal to pi times the | |
us to the fourth power multiplied by the pressure difference,
divided by 8 times the coefficient of viscosity represented by the symbol eta multiplied by the length of the pipe.
So, let's consider point A and B.
L represents the left between those two points of interest.
At point A, we have a pressure 1, point B pressu... | |
, R is the radius of the pipe,
and 8 is the coefficient of viscosity, which is based on the fluid.
So, you have to look that up in the table.
Now, the volume flow rate delta V over delta T, in some textbooks, is represented by the symbol Q.
And keep in mind, the volume flow rate is the area times the flow speed or the | |
city.
In order to apply PoCA's law, some conditions have to be met.
So, first, the fluid has to have laminar flow.
So, what that means is that the fluid has to be traveling straight.
So, it can't be crossing over with another section of the fluid.
It's simply a travel straight.
So, it can't be any turbulent flow.
So, i... | |
n't be moving too fast where it's not traveling straight anymore.
The second is that we need to have an incompressible fluid.
So, what does that mean?
That the fluid is incompressible.
That means that the density of the fluid is constant.
The density doesn't change.
Now, going back to the equation, what we saw that the | |
ume flow rate is piR to the fourth power times the change in pressure P1 minus P2,
divided by 8 times 8 at times L.
We can see that if we increase the radius of the pipe, the flow rate will increase.
Now, if you double the radius of the pipe, the flow rate will increase by 2 to the fourth power.
As we can see, Q is | |
tional to the fourth power of R.
So, if we triple the radius, the flow rate will increase by 3 to the fourth power, which is 81.
So, the volume flow rate is highly dependent on a cross-sectional radius of the pipe.
Now, it also depends on the coefficient of viscosity.
If you increase the coefficient of viscosity, the | |
e flow rate will decrease.
So, for very viscous liquids like honey and syrup, a greater pressure is required in order to keep the fluid at a constant flow rate to keep it moving at a certain speed.
So, the flow rate is inversely related to the coefficient of viscosity, and it's also inversely related to the left of the | |
e,
because that's in the bottom of the equation.
However, it's directly related to the pressure difference.
So, if you increase the pressure difference between 2 points, the volume flow rate will increase.
Now, this information is important when you're dealing with blood vessels, for example.
Let's say, if this is a bl... | |
vessel, and there is, let's say, some plaque buildup, we'll use a different color.
So, let's say, we have cholesterol in your blood arteries.
Let's say that the radius of this blood vessel is 1 centimeter, which is pick a number.
And now, in this area, it's 0.5.
So, the cross-sectional radius has been reduced by a fact... | |
f 2.
So, if the pressure difference is the same, then the volume flow rate is not going to be 1.5 raised to the 4th power,
which means that it's going to be reduced by a factor of 16.
So, if you decrease the cross-sectional radius, the volume flow rate decreases.
And so, in order to maintain a constant volume flow rate... | |
u have to increase the pressure difference by a factor of 16.
And so, that puts more work on your heart.
Which means that the heart has to work a lot harder to maintain the volume flow rate.
It has to increase the pressure difference across this particular blood vessel.
And that's not good.
So, a lot of cholesterol in | |
diet has a negative impact on your heart.
And you can see that with quasi-ase law.
Now, to overcome this, or to basically increase the volume flow rate,
what you could do is increase the temperature.
Whenever you increase the temperature of a fluid, the coefficient of viscosity decreases.
And so, the fluid has less | |
tance towards flow.
And as the coefficient of viscosity decreases, that's inversely way to the volume flow rate,
so the volume flow is going to increase.
So, for example, if you compare cold honey and warm honey, warm honey can flow down
and incline a lot faster than cold honey, because the temperature is higher.
And s... | |
ome people, they always like to have a hot beverage in the morning.
And some folks like it had as possible.
Whenever you can increase the temperature of the fluids in your body,
then you can increase the volume flow rate through the blood vessels.
So, now let's work on this problem.
Water has a coefficient of viscosity... | |
0.001,
Pascal's time seconds at 20 degrees Celsius.
What pressure difference is necessary to keep water moving with a volume flow rate of 0.05 cubic meters per second
through a 3-meter long pipe with a radius of 2 centimeters.
So, let's draw a picture.
So, we have a radius of 2 centimeters,
and the left of the pipe is ... | |
ters.
Let's call this point A and point B.
We need to calculate the pressure difference between these two points.
And the volume flow rate Q is 0.05 cubic meters per second.
So, we know that Q is equal to pi R raised to the fourth power times the pressure difference,
divided by 8, A at a times L.
So, let's isolate delt... | |
So, let's multiply both sides by the denominator of the right side.
So, we have A Q times A at a times L,
and that's equal to pi R to the fourth power times delta P.
So, now let's divide both sides by pi R to the fourth power.
And so, the pressure difference is going to be 8 times the volume flow rate times the | |
t
of viscosity times the left of the pipe divided by pi R to the fourth power.
So, let's plug in the information that we have.
So, the volume flow rate is 0.05,
the coefficient of viscosity is 0.001 at 20 degrees Celsius,
and the left of the pipe is 3 meters.
And then we need to divide it by pi times the radius to the | |
th power.
And the radius is 2 centimeters, which is 0.02 meters.
So, go ahead and type this in your calculator.
So, the pressure difference is 716.02 Pascal's.
So, let's understand what this means.
So, I'm going to round that to 716.
So, let's say if we have a very long pipe.
Let's call this 0.8, 0.B, 0.CD and E.
And t... | |
distance between each of these points, let's say it's 3 meters,
which was the length that we use in the last problem.
And let's say that the pressure at 0.8 is 9,000 Pascal's.
Now, the pressure difference from water flowing in this particular pipe
with across section radius of 2 centimeters.
We said it's 716.02 Pascal'... | |
ut let's say 716.
So, as the water flows, the pressure is going to decrease by 716 Pascal's.
If it's flowing in this direction.
If it's flowing in the opposite direction, then it's going to like increase towards the right.
But in the direction that it's flowing, it's going to decrease in that direction.
So, at 0.B, the | |
ssure is going to be 9,000 minus 716.
So, every 3 meters, the pressure is going to decrease by 716.
So, at 0.C, the pressure is now 7,568 Pascal's.
And at 0.D, it's 6,852 Pascal's.
And at 0.E, it's going to decrease even further.
Until it comes to a stop.
So, friction, which opposes motionless call friction F.
Notice t... | |
because of friction, the pressure is decreasing along the pipe.
Now, let's move on to number 2.
Engine oil with a coefficient of viscosity of 0.2 Pascal's time seconds.
Flows through a 5-meter long pipe with a radius of 5 cm and a pressure difference of 95,000 Pascal's.
Calculate the volume flow rate and mass flow rate... | |
this pipe.
So, let's draw a picture.
So, we have a 5-meter long pipe.
We're at least 5 meters between the 2.1s of interest.
And the cross-sectional radius is 5 centimeters.
We have a pressure difference of 95,000 Pascal's.
To find a volume flow rate, it's going to be pi times the radius to the 4th power.
Times the | |
ce in pressure divided by 8 times 8 times 0.
So, it's pi times the radius, which is 5 cm divided by 100 or 0.05 meters.
The pressure difference is 95,000 Pascal's.
And the coefficient of viscosity is 0.2 and the left of the pipe is 5 meters.
So, we just got to plug these numbers in.
So, the volume flow rate is 0.233 | |
cubic meters per second.
So, now that we have the volume flow rate, how can we calculate the mass flow rate?
Mass is density times volume.
So, the mass flow rate is going to be the density of the fluid times the volume flow rate.
The density is 720 kg per cubic meter.
And the volume flow rate is 0.233 cubic meters per | |
cond.
So, the unit cubic meters will cancel.
So, let's take our answer and multiply it by 720.
And so, this is going to be about 168 kilograms per second.
So, that's the mass flow rate of the fluid in this pipe.
Number 3, a pressure difference of 50,000 Pascal's is needed to maintain a constant volume flow rate.
Of 0.0... | |
ubic meters per second.
Calculate the power required to achieve this.
So, what should we do in this problem?
So, what is the formula for power?
You know that power is the ratio between work and time.
And work is forced times displacement.
Now, what I'm going to do is multiply the top and the bottom by the area.
A over ... | |
1. So, I can do that.
Whenever you multiply something by 1, the value of that something does not change.
Now, what I'm going to do is take A and move it to the bottom of F.
So, power is forced divided by area times displacement times area over T.
Now, volume is area times height.
Area is in square meters height is in | |
rs and volume is in cubic meters.
And the height is basically the vertical displacement.
So, we can say that volume is area times displacement.
So, I'm going to replace this with V.
And force over area is pressure.
So, have pressure times the volume divided by T.
Delta V over Delta T is the volume flow rate, which is | |
sented by the symbol Q.
So, power is pressure times the volume flow rate.
Now, we have the pressure is 50,000 pass gals.
And we have the volume flow rate.
So, the pressure difference is 50,000, which is Newton's per square meter.
That's one pass gals.
And the volume flow rate, that's 0.025 cubic meters per second.
So, ... | |
power is going to be 1250 watts.
The unit that we have left over is Newton's times meters per second.
And the unit times meters force times displacement.
That's the joule.
And so, it's joules of a second, which is a walk. | |
In today's video, I want to give you a list of formulas associated with spontaneity.
So this is going to include entropy, enthalpy, and Gibbs free energy.
So here's the first one.
The entropy change delta S of the universe is the sum of the entropy change of the system
plus the entropy change of the surroundings.
Now y... | |
te the entropy change of the surroundings by dividing the enthalpy
delta H over the temperature.
By the way, this equation is associated with the second law of thermodynamics.
Now it's good to know that the entropy, or more specifically the positional entropy of a gas,
is greater than that for a liquid and a positional... | |
iquid is greater than that of a solid.
By the way, whenever you increase the temperature of a substance, the entropy of that substance will typically increase.
Now the next formula you need to be familiar with is free energy change, or Gibbs free energy, delta G.
Delta G is equal to the enthalpy minus the temperature t... | |
e in entropy.
Now when using this formula, you need to pay attention to the units because,
specifically, you'll be given the enthalpy in kilojoules per mole, but the entropy is usually given to you in joules per mole per Kelvin.
So first you need to decide, do you want your delta G value to be in kilojoules per mole or... | |
ole?
Typically this usually is reported in kilojoules per mole, so what that means is you need to convert entropy from joules to kilojoules by dividing it by 1000.
When you have that, then the formula is going to work.
So just be mindful of that whenever you use that formula.
Now if you need to calculate the non-standa... | |
ue from the standard free energy change value,
if you see a circle on top, it just indicates that you're dealing with the standard free energy change.
So non-standard delta G is equal to the standard delta G plus RT.
That length Q, Q is the reaction quotient.
So if you have some reaction to A plus 3B turns into 4C plus... | |
hat all of this is in either the aqueous phase or the gas phase,
then Q is going to be C raised to the fourth power, D raised to the fifth power over A squared times B to the third.
So just like equilibrium, the reaction quotient is equal to the products over the reactants where the coefficients become exponents.
Now t... | |
lates delta G to the equilibrium constant K.
The standard free energy change is equal to negative RT, LNK, where K is the equilibrium constant.
And you can calculate K from this equation.
It's E raised to the negative delta G over RT.
And for all of these formulas, R is the energy constant 8.3145 joules per mole per Ke... | |
times you need to calculate the standard free energy change for reaction.
To do that, you need to take the sum of these standard free energy change,
or rather the standard free energy of formation of the products multiplied by the coefficients in the balance reaction.
And then minus the sum of the free energy change st... | |
rgy change of formation of the reactants.
So there's a lot of different delta Gs out there.
So just review, if you see G, G is Gibbs free energy.
Delta G is the change in free energy, or change in Gibbs free energy.
If you see Delta G not, that's the standard free energy change.
So this is free energy change, free ener... | |
ee energy change.
Now if you see G of formation, this is the free energy of formation.
So that's basically, if you would have had a reaction where you're forming a substance from its constituent elements in our natural standard states,
you'll get this value for the substance that you're considering.
And you could look ... | |
in a table.
Now I do have some example problems that will show you how to use this formula to calculate the standard free energy change of a reaction.
But I wanted to go over the difference between all of these symbols.
So if you see the F, it just stands for a formation.
Now you can calculate the enthalpy change of a ... | |
a similar formula.
So it's the sum of the standard enthalpy's information for the products minus the sum of the standard enthalpy's information for the reactants.
So I'll just put R for reactants.
Now you can also calculate the entropy change of a reaction in the same format.
And again, I have example problems which I'... | |
st in the description section below, so you could see how to use these formulas.
But typically, if you have the thermal dynamic table data, you would typically use that data to find delta H and delta S and using that.
You can calculate delta G.
But you could use the table to get delta G of the reaction as well.
Now, if... | |
atural log of the equilibrium constant with the reciprocal of temperature, you can get a straight line plot.
And the formula that describes that linear equation is this.
The natural log of the equilibrium constant is equal to delta H over R times 1 over T plus delta S over R.
So notice that this linear equation is in s... | |
form.
So Y is L and K, so you will plot this on a Y axis.
X is 1 over T, so you would plot that on the X axis.
Negative delta H over R is the slope and delta S over R correlates to the Y intercept.
So if you have a straight line plot of L and K versus 1 of T, you can get the enthalpy and the entropy from that plot.
So ... | |
mine the slope of the line, the enthalpy will be negative times the slope times R where R is 8.3145 jules per mole per Kelvin.
Once you know the Y intercept, you can get the entropy change by multiplying the Y intercept by R.
And then once you have both delta H and delta S, you could use this formula to get delta G.
So... | |
additional formulas that you want to take down for notes.
In addition to that, whenever the standard free energy change is less than 0 or when it's negative, what you have is a spontaneous reaction.
And when delta G is negative, K is greater than 1.
And that's based on this equation, delta G is equal to negative RT at ... | |
hen K is greater than 1, L and K will be a positive number, but because we have a negative sign here, delta G will be negative.
So for a product favored spontaneous reaction, K is typically a large number, like Stimphky larger than 1,
and for a spontaneous process, delta G is negative, which means that it's also less t... | |
delta G, the standard free energy change is equal to 0, based on this equation, K is going to equal 1, because L and 1 is 0.
And when delta G is 0, the reaction is at equilibrium.
Now when the standard free energy change is greater than 0, that means that it's positive, and it also means that K is less than 1.
Now K is... | |
ve, so when K is less than 1, you need to understand that it's between 0 and 1.
When delta G is positive, or when K is very small, what you have is a non-spontaneous reaction, that is, it's non-spontaneous in the 4 direction, which means it's spontaneous in reverse direction.
So in the 4 direction, it's non-spontaneous... | |
d favored, which means the reaction will prefer to go to the left as opposed to the right.
So those are the most common equations that you're going to encounter if you're studying sponson 80, that is, if you're studying like entropy, enthalpy, and Gibbs free energy.
By the way, for those of you who want to print out of... | |
and more, I feel free to check out the formula sheet, which I'm going to post down in the description section below. | |
So this video is an introduction into circles.
The radius of a circle is a segment that connects the center of the circle, which will
call A to any point on a circle.
So segment AB represents the radius of a circle.
The diameter passes through the center of the circle and is twice the length of | |
he radius.
So this is the diameter, and let's call this A, B, C.
So segment AB, C, is the diameter of a circle.
Now on the left, the center of the circle is point A, so you can describe the circle
like this.
On the right, you can describe the circle based on the center point B.
Now the area of a | |
ircle is pi r squared, and the circumference of a circle is 2 pi r.
There's the circumference of a circle is basically the perimeter of a circle.
It's the distance around the circle.
Now the diameter is twice the radius, so we can replace 2 r with the diameter.
So you can also calculate this | |
umference using this equation.
This is pi times the diameter.
Now sometimes you may need to calculate only a section of the circle.
For example, let's say, this is point A, B, and C.
And let's say B is the center of the circle.
And we wish to calculate the area of this sector.
How can we do so? | |
d also how can we calculate the left of this arc, which I'm going to call this?
So if B is the center of the circle, that means segment A, B, and segment B, C represents
the radius of the circle.
Now to calculate the shaded region, the area of the shaded region rather, it's very similar
to the | |
ea of a circle, but with some adjustments.
Notice that we have a fraction of the circle.
So whatever this angle is, theta, indigrees, it's going to be theta divided by 360, because
the entire circle has an angle of 360.
So theta over 360 represents the fraction of the circle multiplied by the | |
of the circle.
So this equation will give you the area of a sector, or the shaded region that we have there.
Now to calculate the arc length of that sector, it's basically a fraction of this a conference.
So it's theta divided by 360 times this a conference of the circle.
So that's how you can | |
alculate the arc length.
So as the arc between A and B, you can describe it this way.
So this represents the arc between points A and point C, and the measure of arc A to
C, that's an angle.
So in this example, it's theta.
So theta represents the measure of arc A C.
To calculate the length of arc | |
A C, you can use this formula.
Now let's talk about chords.
Let me draw a better circle.
It's not really that much better.
Now what exactly is a chord?
A chord is a line segment that connects two points that are on the edge of the circle.
So A B is a chord.
C D is also a chord.
Now let's say | |
this point is a center.
Let's call this D E F.
So we have circle E, or with center E, and D F is a chord.
Now a chord that passes through the center of a circle is also known as the diameter of
the circle.
So D F is basically a chord, and it's the diameter of the circle, which means E F | |
s
the radius of the circle, and D E is also the radius of a circle.
So a chord is simply a line segment that starts at one end of the circle and touches another
point on a circle.
Now what happens if we form an angle using two chords?
So the two chords are chord A B and chord B C.
Notice that | |
s angle angle A B C, which is known as an inscribed angle, it has an intercepted
arc A C, which we can write like this.
Now let's say if the inscribed angle is 50 degrees.
So if angle A B C is equal to 50 degrees, what do you think the measure of arc A C is?
It turns out that the intercepted arc | |
as an angle that's twice the value of the inscribed
angle, so this is going to be 100 degrees.
So make sure you understand that.
Now let's say if you have an angle that touches the center of a circle as opposed to
a point on a circle.
So let's say C is the center, and this is 40 degrees.
And | |
s call this point A and point B.
What is the measure of arc A B?
The measure of arc A B is the same as the angle that touches the center of the circle.
So it's going to be 40 degrees.
But if you have an inscribed angle, let's say this is 80, the inscribed angle is half of the
intercepted arc, so | |
his is going to be 40.
Now let's work on some problems.
So we're given the diameter of a circle, and it's 8 centimeters.
Our goal is to calculate this a conference and the area of a circle.
So let's draw a picture.
So this is the diameter, this is 8 centimeters.
This a conference is simply pi | |
es the diameter, so it's pi times 8, or you can write it
as 8 pi.
So that's the exact answer of this conference.
Now if you want to get the decimal value for that, just type in 8 pi in your calculator.
So you should get 25.1 centimeters as this conference.
So that's a rounded answer.
Now the area | |
is pi r squared.
So first we need to calculate the radius.
The radius is 1 half of the diameter, so it's half of 8, which is 4 centimeters.
So the area is going to be pi times 4 squared, or 16 pi.
So 16 pi is the exact answer, and the units for that are square centimeters.
This is in centimeters. | |
And 16 pi is 50.27 square centimeters as a decimal, approximately.
Number two, the area of a circle is 81 pi.
What is this a conference of the circle?
So we know the area is pi r squared, and the area is 81 pi.
So let's calculate the value of r.
So first we could divide both sides by pi.
If we | |
so, these will cancel.
So 81 is equal to r squared, and now we just got to take the square root of both sides.
The square root of 81 is 9, and so that's the radius.
It's 9 units long.
This a conference is 2 pi r, so it's 2 pi times 9, which is 18 pi units.
So as a decimal, that's 56.55 units | |
g.
Now the diameter is twice the value of the radius, so it's 2 times 9, or 18.
And so that's the length of the diameter of a circle.
Number three, the radius of the circle below is 7 inches.
What is the angle measure of arc AC?
So this is arc AC.
So the angle measure of arc AC is equal to this | |
gle, where B is the center of the circle.
So Bc is 7 inches long, and A, B is 7 inches long.
So the measure of the arc is 150.
Now what is the length of arc AC?
So to calculate the arc length, which you can do no SS, it's equal to theta divided by
360 times this is a conference of a circle, | |
is 2 pi r.
So this angle is 150, we need to divide that by 360.
And then multiply that by 2 pi times the radius, which is 7.
So 150 divided by 360 times 2 times 7.
This turns out to be 35 pi over 6.
Now if you want to see how to get that answer without a specialized calculator, first
we can | |
l the zero.
And 15 we can reduce that to 5 times 3.
36, that's 3 times 12.
And 12, I'm going to break that into 6 times 2, and we have 2 and 7 on the outside.
So notice I can cancel a 3, and I can cancel a 2.
So I'm left over with 5 times 7, which gives me 35, and I have a 6 on the bottom.
So | |
t's where the 6 comes from, and plus we have pi.
So the arc length, so arc AC has a length of 35 pi over 6.
Now as a decimal, this is 18.3.
Now let's calculate the area of the shaded region.
So the area of the blue shaded region is going to be the angle divided by 360 times the
area of the entire | |
circle.
So the angle is 150, and we're going to divide that by 360, and the radius is 7, so we're
going to multiply it by 7 squared.
So once again we can cancel a zero.
15 is 5 times 3.
36 is 12 times 3.
And 7 squared is 49.
So 49 times 5 is 245.
So this is going to be 245 pi divided by 12.
So | |
at's the exact answer.
And as the decimal, this is about 64.1 square units.
So this represents the area of the sector.
So calculate the value of x and y in each figure shown below.
So for the circle on the top left, we can see that b is the center of the circle.
Therefore x and the measure of arc | |
AC has to be the same.
So for the first problem, x is equal to 80 degrees.
Now for the second problem on the top right, the inscribed angle is going to be half of the
intercepted arc.
So y is one half of 60.
So in this case it's 30 degrees.
Now for this one, the intercepted arc is twice the | |
of the inscribed angle.
So x is going to be twice the value of 80.
So it's two times 80 or 160 degrees.
Now what about the third problem on the bottom left?
One thing I forgot to do is tell you what we're trying to calculate.
So let me do that now.
What is the value of x in this figure?
Where x | |
epresents this angle right here?
So notice that it intercepts these two points, which we can call bc.
And notice that a is the center of the circle.
So therefore, chord bc is the diameter, which by-sex circle is two parts.
Therefore half of 360 is 180.
So the intercepted arc is 180.
Which means | |
hat the inscribed angle is half of the intercepted arc.
So x is going to be half of 180, which means that it's 90.
So this triangle is a right triangle.
Number five, if a b is 48 and bc is 14, what is the area of the shaded region?
So the area of the shaded region is the difference, a-s we | |
nt the area of the shaded region,
is the difference between the area of the circle, a-c minus the area of the triangle, a-t.
Now the area of the circle is pi r squared.
The area of a triangle, particularly a right triangle, it's one half, base time type.
So triangle a-bc, which looks like this, | |
'm going to draw it this way.
So this is the hyponus.
So this is vertex b, this is a, and this is c.
Now a-b is 48, and bc is 14.
Now notice that a-c is the equivalent of the diameter of the circle.
If we can calculate the diameter, we can calculate the radius.
Now bc is the base of the triangle. | |
So bc is 14.
So low case being this equation is 14.
The height of the triangle is 48.
That's a-b.
So now we just got to calculate the radius.
So let's focus on calculating the diameter first.
So for a right triangle, we can use the Pythagorean theorem.
C squared is equal to a squared plus b | |
ed.
C is the hyponus, which is across the box.
So that's the same as the diameter.
A, it's going to be 14.
You can make a or b-14.
So if a is 14, b has to be 48.
Now 14 squared is 196.
48 times 48 is 234.
So if we add up 2304 plus 196, that will give us
2500, and that's equal to d squared.
So to | |
alculate d, we got to take the square root of both sides.
2500 is the same as 25 times 100.
The square root of 25 is 5, and the square root of 100 is 10.
So 5 times 10 is 50.
So the diameter is 50.
The radius is 1 half of the diameter.
So it's half of 50, which is 25.
So now we have everything | |
at we need to calculate the area
of the shader region.
So it's going to be pi times 25 squared.
So 25 squared is 625.
Half of 14 is 7, and 7 times 48 is 336.
So this is the exact answer.
It's 625 pi minus 336.
Now let's get the answer as a decimal.
625 pi is 1,963.5.
If we subtracted by 336, this | |
is equal to 627.5.
And so this is the area of the shader region. | |
I've just written down the equation.
Live in a null, we can get pretty much everything we need just off of this.
What are the first things that are asked or lost?
It says write down it's center amplitude and extremes of displacement.
That's a bit of a weird phrase that we can get to,
you might be able to infer what tha... | |
's going up and down and the middle of it is just x equals zero, right?
So when we introduce this kind of vertical shift,
whether it's up or down, you just change where that middle is.
So which part of our equation is that?
That's the three, right on the end there.
So I'm just going to say center of motion.
And then I'... | |
I'm going to say it's x equals three.
Because that's what displacement is.
It's x, choose the right, perinary,
you're going to have a lot flying around, okay?
Next thing, amplitude.
You didn't even need to know that this is simple harmonic motion to look at this function and tell me which number,
which value is going t... | |
se
a because we're borrowing about that as equal to two.
Now the extremes of displacement, what does that mean?
Well, extremes are like what are the furthest places you can go?
What's the most extreme value?
So in this case, and some people will call them extremities.
Now I'd like to use that because it's just lazy, it... | |
t you can go from x equals three?
You can go two units down to one and then what's the highest you can go from three?
Up to five.
So I can say my extremities are x equals one and x equals five.
That's it.
I just rid them off.
It's my a plus or minus.
Sorry, it's my x plus or minus a.
Okay.
And is the furthest place?
Ah... |
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