Monday, April 16, 2018

Finding Time Using the Distance Equation

Any calculation of time (t) is relatively easy in the cases where either vi or a was zero. In the cases where neither are zero, the math results in a second degree polynomial equation such as:
0 = x2 + 3x - 12
Where this occurs in physics of motion is in the full distance equation:
df = di + (vi)(t) + (1/2)(a)(t2)
Although this does not exactly match the expected form for a quadratic equation, it can easily be rearranged as such:

df = di + (vi)(t) + (1/2)(a)(t2)
0 = di - df + (vi)(t) + (1/2)(a)(t2)
0 = (1/2)(a)(t2) + (vi)(t) + di - df

Keep in mind that di - df will yield a number when the values are plugged in and simplified.
To solve these problems, the steps are the same for any problem in science, BUT the algebra becomes harder.
After you have plugged in the numbers, combine like terms and simplify. Suppose the following:
df = 89
di = 50
vi = 4
a = 6
and you need to find t
df = di + (vi)(t) + (1/2)(a)(t2)
89 = 50 + (4)(t) + 1/2(6)(t2)
0 = -39 + 4t + 3t2
(Put it in normal quadratic form.)
0 = 3t2 + 4t - 39
Now you can either factor or use the quadratic formula to find the values for t:

0 = ( t  - 3 )( 3t + 13)
0 = t - 3    AND   0 = 3t + 13
3 = t        AND   -13/3 = t

Since time cannot be negative within the context of classical physics, only t = 3 is a valid answer.
Using the quadratic equation will yield the same results.

While it is far easier to find t when either a or vi is zero, the math to find t when that is not the case is not beyond the skills of a student taking an introductory physics class.
For another look at this process, check out this video:


Thursday, March 22, 2018

Physics of Motion: What is What?

Physics of Motion: What is What?


The equations:


df = di + (vi)(t) + (½)(a)(t2)   This will answer almost all “how far” questions.
Use this if:
Only 1 velocity is given
You have a distance given (df is rarely 0, but sometimes it is the unknown)


vf = vi + (a)(t)    This will answer almost all “how fast” questions.
Use this if:
2 velocities are given
You don’t have distance


v(ave) = (vi + vf)/2
Use this if:
You need to find the average velocity
You know BOTH vi and vf (you might have to calculate vf!


vf = (vi + ∆v)
Use this if:
You need to find the final velocity AND
You know BOTH vi and how much v changes (e.g. speed increases by 10 m/s)


df = (di + ∆d)
Use this if:
You need to find the final distance, total distance, or final position AND
You know BOTH di and how much d changes (e.g. it moves 10 meters)


The hints and helps:


It is vi if…
“…traveling at a rate of…”
“…moving at…”
"…has a velocity of…"


If it is “at rest” then vi = 0 and di is probably 0

The words "begin" and "starts" generally go with the initial values.


If it is “moving at a constant” rate or if it “has a constant velocity” then a = 0

It is possible that something not given (but which is not the thing to be found) should have a value of 0...


Monday, October 23, 2017

Summary of Atomic Theory

The 21st Century understanding of atomic theory is very effective in helping explain and predict how substances interact. It is the result of a growing body of knowledge that dates back centuries.

Democritus

In the middle of the 5th Century BCE, Democritus proposed ideas that were correct in many ways. Democritus believed that all matter consisted of extremely small particles that could not be divided. He called these particles atoms from the greek word transliterated as atomos, which means "uncut" or "indivisible." 
He thought there were different types of atoms with specific sets of properties for the different substances in creation. The atoms in liquids, for example, were round and smooth, but the atoms in solids were rough and prickly. 
Though Democritus's ideas didn’t catch on, he was surprisingly accurate in his assumptions. Scientist of the 21st Century eventually concluded that many of Democritus’s ideas were correct, and they extended the understanding of atomic theory in stages that led to the current model.

John Dalton

Dalton proposed the theory that all matter is made up of individual particles called atoms, and in it, he identified several things that have lasted into the modern atomic theory:
  • All elements are composed of atoms.
  • All atoms of the same element have the same mass, and atoms of different elements have different masses.
  • Compounds contain atoms of more than one element.
  • In a particular compound, atoms of different elements always combine in the same ratios.



J. J Thompson
Thompson's experiments provided the first evidence that atoms are made of even smaller particles that have positive and negative charges.


Ernest Rutherford
Rutherford's model extended what was previously understood by identifying that atoms have a central, relatively dense (compared to the entire volume of an atom) nucleus around which electrons move. 

Niels Bohr

  • In Bohr's model, electrons move with constant speed in fixed orbitals around the nucleus.
  • Electrons must orbit the nucleus in one of several fixed, specific orbits, and each orbit represents a specific energy level.
  • The first orbital represents the lowest electron energy level, and the other orbitals represent progressively higher and higher energy levels.


Electron Cloud Model

  • Evidence following Bohr's work led to the understanding that the electrons do not orbit the nucleus like a planet.
  • While they do exist at specific energy levels and occupy orbitals, their position in the orbital is never 100% certain. They are somewhere in the orbital, but exactly where cannot be known specifically. Each orbital can be, therefore, conceived as an electron cloud. 


The progression of Atomic Theory in the modern era took place in various steps and stages. The resulting body of understanding results in a model of atoms that is highly effective and useful in understanding and predicting the behavior of substances.


Thursday, October 5, 2017

Solving Science Word Problems (and any other kind of word problem, too.)

Throughout science, there inevitably comes a time when it is necessary to use some "known relationship" to figure out what happened or what might happen. In physics and chemistry, especially, this is the case, but sciences like sociology use "known relationships," such as population growth models, to explain or predict certain happenings.

In many, many cases, the "known relationship" is expressed as a mathematical equation. The relationship between the rate something happens and the amount of something produced is very common, so this discussion will use a couple of rate problems as its example.

In most general terms, the amount of something produced is equal to the rate of production times the amount of time production took place. For instance, in a displacement problem, rate is how far something moves in a given time and motion of an object is produced. In a general form a rate problem might look like this: 
O = R•t 
where O is the output, R is the rate, and t is the time.

So what is this process of solving science problem that will use rate problems as its example?

There are three steps to solving ANY science problem (or word problem of any kind, for that matter).

The following two examples will be used as the steps are discussed:

EXAMPLE 1: A baker's oven will hold only 1 pan of cookies, and each pain has space for 12 cookies. The baking time on cookies (including putting the dough on the pan) is 20 minutes. Therefore, the rate cookies are baked is 12 cookies per 20 minutes. How many cookies can be baked in 80 minutes? 
EXAMPLE 2: A car travels at an average rate of 20 MPH for 3 hours. How far does it go.

The first step is to write down what is given AND what is asked.

Example 1:
R = 12 cookies/20minutes
t = 80 minutes
LOOKING FOR "How many cookies"
Example 2:
R= 20 MPH
t=3 hours
LOOKING FOR "How far does it go"
A student familiar with physics motion problems would see Example 2 as a "distance" problem and would use the distance equation variables. In such a case, R would be velocity, v, t would be time, t, and LOOKING FOR would be distance, d (or in some cases, s). 
The second step is to write the equation that relates to the question.

Example 1:
Cookies = R•t

Example 2:
d = vt

From time to time, it is necessary to rearrange an equation so that it yields the answer you are trying to find. For example, if you trying to find time in Example 2 above, you will need to isolate the t variable. MANY TIMES it is easier to do the algebra BEFORE you plug in the numbers and units!

d = vt               (To find t divide both sides by v.) 
d/v = vt/v           (The v on the right side cancels.)
d/v = t
Once you have an equation in the form you need, you are ready for the next and final step.

The third step is to plug in what was given and solve. Assuming the second step was completed, the following solutions would emerge:


Example 1:
Cookies = R•t
Cookies = 12 Cookie/20 min • 80 min
Cookies = 960 Cookie•min / 20 min
Cookies = 48 Cookie 

Example 2:
d = vt
d
=  20 mi/hr * 3hr
d = 60 (mi•hr)/hr
d = 60 mi

For additional examples and a video explanation, check this out:





SUMMARY

There are three steps to solving problems.

1: Write down what is given.
2: Write down the relevant equation.
2b: Rearrange the equation so that it yields the answer you want.
3. Plug in and solve.


Tuesday, August 1, 2017

Physical Quantities: Equivalency and Conversion

Objects in the physical world can be described according to various physical properties, such as length, volume, mass (or weight). Since the beginning of humanity, people have come up with ways to measure things in a standard way.

Standards and Units

Measurements were usually created to compare to some reference object or other agreed-upon standard. For instance, if trading sea shells, the standard would be… a seashell. One sea shell equaled one seashell.

Okay… hang in there… keep going…

So, along the way, magic occurred (not really) and people began to equate the word "one" with "unit." So, if you said, "Give me eight units of seashells," since seashells were measured in… themselves… you would get eight seashells.

But suppose a guy sold sand. Selling grains of sand would be… dumb. Suppose (sticking to the beach motif) he had a coconut hollowed out. His thing was, one scoop of sand for one seashell…

So for sand, the unit would be scoops.

A buyer would say, "Give me eight scoops of sand," and would pay eight seashells for it.

It must be time for a definition!

Unita quantity chosen as a standard in terms of which other quantities may be expressed. (Oxford dictionary.

Not very helpful, but in the right direction!

Anne Marie Helmenstine, Ph.D. (https://www.thoughtco.com/definition-of-unit-in-chemistry-605934) defines it this way:

UNIT DEFINITION
A unit is any standard used for comparison in measurements.

Going back to strictly intuition, most people already know what units are. "Comparison" is a good word to hang onto, so do that.

Now think about things you already know. Gallons of gasoline. Pounds of lunch meat. Miles to the next town.

Using a standard unit like miles allows us to make reasonable comparisons. A mile is a mile is a mile. So if it is 200 miles to Townville and 400 miles to Villaton, since a mile is a mile is a mile, it must be twice as far to Villaton.

If I have 4 gallons of paint and you have 2, I have twice as much paint, since a gallon is a gallon is a gallon.

So, think about things around you. What are the standard units for them.

Distances between towns?
Shampoo?
Foundation?
Firewood?
Soda (pop)?

Chances are, for at least one of those, you thought of different standards. For example, you might find 20 oz bottles of soda as well as 2 liter bottles!

There are times when converting from one standard unit to the other is desirable.

Conversion and Equivalency

The process of converting from one standard unit to another boils down to finding how many of one thing is equal to one thing of the other.

Good news, #1: The math on this is easy.

Good news, #2: You can usually look it up on Google.

Bad news, #1: An explanation follows anyway.


To start with a definition of conversion…

Conversion: The process of finding out how many units from one standard are equal to how many units from another standard.

Suppose you discover that, in a fantasy novel, the people in one town sell milk by the Nallog and in another by the Ecnuo. Some guy has a barrel on which are etched lines for both Nallogs and Ecnuos, and another dude sees that the milk comes to the 2 Nallog mark and also the 256 Ecnou mark.

In the (silly) example, you can see that 2 Nallogs = 256 Ecnous. That is, the two quantities are equivalent.

Finding out how many Ecnous are in a Nallog is easy! Divide the bigger number by the smaller. Bam!

256 ÷ 2 = 128

So 1 Nallog  (the bigger unit) is equivalent to 128 Ecnou (the smaller unit).

In the (silly) example above, the conversion factor is 2.

Conversion factor: A number that, when multiplied will convert one unit to another.

Oxford Dictionary says it like this: an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

To find (or derive) a conversion factor for anything, all that is needed is to know how much of something is present in the two different units. Then, to find the conversion factor, divide the larger by the smaller of the quantities. The quotient (number after you push the equal button) will be how many of the smaller thing (the bigger number) are in the bigger thing (smaller number).

No, that's not confusing, is it?

EXAMPLE

A container has 591.5 milliliters of shampoo in it. The label also says it has 10 clarkens (which are cleverly abbreviated as "Clar"). Find the conversion factor for milliliters and clarkens.

591.5 ml
_______  =   59.15 ml/Clar
10 Clar


Thus, 1 Clar is equal to 59.15 ml.

BAM! Easy math!



Conclusion

It is always best to just measure things in the units you need. But sometimes, that's impossible. Sometimes your most accurate measuring device provides units that you have to convert.

A solid understanding of equivalency and conversion prepares students to face the demands of science and engineering.


Monday, July 31, 2017

Physical Quantities: Quantities and Units

When looking at the world around, people pretty much automatically attend to physical quantities, pretty much without thinking about them. The youngest child intuitively can judge differences with some skill.

By the time the child reaches school age, the following conversation would seem drastically out of place:

Teacher: "Mary, how far can you run?"
Mary: "I can run six pounds! Very heavy!"

Almost all children would know that far and heavy measure different physical quantities.

As most readers of this would know, one of the jobs of science is to complicate things. No… not that… To clarify things by assigning specific words with specific meanings in order to make discussion precise and accurate (which are both words surrounded by confusion, ironically).

Therefore, with regard to physical quantities, a certain set of words are used to describe specific things.

Included in those sets of words are two sides to every quantity. One side is the concept of what is being described. Mary, in the dialog above, gave an answer about one quantity with words used for a different. The other side of the concept is, for every quantity, specific ways that it is measured (which are called units). Pounds do not go with a "how far" question!

First what are the things being described?

Types of Physical Quantities

The following list is certainly not all-inclusive. It includes a few very common and familiar quantities that are measured.

Distance: Distance is a measure of… okay, distance is so common it actually has different meanings, so in science, there are different words that are used so in order to be more specific.
Distance (case 1): The measure of how far two points are from each other, as in, "The tip of the antenna was 12 meters from the the surface of the window."
Distance (case 2): The length of a path that a moving body takes, as in, "The dog ran from tree to tree through the park until it was at the tree next to the one where it started, covering a distance of 250 yards." Compare with Displacement, next!
DisplacementThe measure in a straight line between where a moving body begins and ends, as in, "The dog ran from tree to tree through the park until it was at the tree next to the one where it started, resulting in a displacement of only 4 yards."
Length, Width, Height: Pretty much what you expect, on this! They measure of how far specific points of an object are from other points on the object.
Other things that would be included in the distance concept are circumference, radius, perimeter, range, altitude, depth… You can probably think of others. But will you?
Volume: (Not talking about sound, which borrowed from this concept for its own purposes!) This also is pretty much what you expect!

Volume is the space in three dimensions that an object or substance takes up (or holds). If a tank holds 25 gallons of gasoline, then the volume of the gasoline that filled the tank would be (duh) 25 gallons. A bottle of soda holds 20 ounces. A different bottle holds 2 liters.

Mass: Mass is a physical quantity that is the result of how many protons, electrons, and neutrons are all in a given space. It is actually not directly observable, though there are devices (scales and balances) that combine with gravity (or other acceleration) and other laws of physics so that it can be measured. Mass is a measure of the amount of matter an object contains.

If two things with the same volume have different masses, one would feel heavier than the other.

Good news: There are scales that measure mass, so you can just plop things down and get a number!

Weight: Weight is a measure of mass under a particular condition. Many people have heard things like, "Well, on the moon, I'd only weigh 12 pounds."

Weight is a basic concept that people are very familiar with. Weight is the degree of heaviness something has.

Time: This is something people very intuitively understand, but which is actually a very abstract concept. Stop reading and write down a definition of time. Not how time is measured! Try a definition of time that does not use seconds, minutes, hours, days, etc. and see what you come up with!

According to the Oxford dictionary, time is the indefinite continued progress of existence and events in the past, present, and future regarded as a whole. 

Time is a very basic concept in science, and fortunately, our intuitive understanding is enough for us to use it.

How do you measure these quantities?

The list below will connect the quantities above with the SOME OF THE units used to measure them. Common units in science are in bold.

Distance: inches, feet, yards, miles, meters, kilometers, centimeters, lightyears

Volume: gallon, ounce, cup, teaspoon, liters, milliliters

Mass: slugs, grams, kilograms

Weight: pounds, newtons

Tuesday, April 25, 2017

Formula Quick Look

The following page is a list of formulas and very brief explanations.

Density is the ratio of a substance's mass to its volume and can be expressed mathematically as


D=M/V
where D is density, M is mass, and V is volume.

Example:

What is the density of an object having a mass of 8 kg and a volume of 2 cubic meters?
D = M/V
D = 8/2
D = 4 kg/m3


Temperature:
In science, we will use Celsius or Kelvin temperature scales to describe temperature.
To convert between Celsius and Kelvin:

Kelvin = Celsius + 273.15
Celsius = Kelvin - 273.15

To convert between Celsius and Fahrenheit:

Fahrenheit = Celsius * (9/5) + 32
Celsius = (Fahrenheit - 32) * 5/9

Gas Laws:

Charles's Law
The volume of a gas is directly proportional to its temperature in kelvins if the pressure and number of molecules are constant.


V1T1=V2T2  

Boyle's Law
The volume of a gas is inversely proportional to its pressure if the temperature and the number of molecules are constant.


P1V1=P2T2

Combined Gas Law
Pressure is inversely proportional to volume, or higher volume equals lower pressure. Pressure is directly proportional to temperature, or higher temperature equals higher pressure.


(P1V1)/T1=(P2V2)/T2 
Example:
If a sample of gas initially has a pressure of 2 atm, a volume of 3 liters, and a temperature of 300 K, what would its final volume be if the pressure changed to 1.5 atm and the temperature changed to 290 K? 

(P1V1)/T1=(P2V2)/T(2 • 3)/300 = (1.5 • V)/290
6/300 = 1.5 V/290
290 • (6/300) = 1.5 V
5.8 = 1.5 V
5.8/1.5 = V
3.87 l = V



Ideal Gas Law

When the number of molecules are included in calculations, the following formula can be used:
PV = nRT

This law can be converted into the following form which will allow memorization of only 1 formula for gas law problems:

P1V    P2V2
_____ = _____
n1T1       n2T





Motion:

Finding final velocity:
vf = vi + at
where vf is final velocity, vi is initial velocity, a is acceleration, and t is elapsed time.

Example:
An object is moving at a rate of 3 m/s and accelerates at a rate of 2 (m/s)/s for 5 seconds. What is its final velocity? 
vf = vi + at
vf = 3 + 2 • 5
vf = 3 + 10
vf = 13 m/s

Finding average velocity:
v(ave) = (vf+vi)/2
where v(ave) is average velocity, vf is final velocity, and vi is initial velocity.


df = di + vit + 1/2at2         
where df is the final, total displacement… 
di is the initial displacement. (How far from whatever point of reference is the object when the thing starts accelerating?)…
 vi is the initial velocity of the object at the beginning of the acceleration.
t is the elapsed time from the beginning of the acceleration until the end of the period being observed. 
(vit accounts for the motion of the object based on its starting velocity. It keeps covering distance at the initial rate, and additionally, it accelerates and covers more distance.) 
a is the acceleration and t is elapsed time.


Example:
An object begins 10 meters from a mark on a track with an initial velocity of 3 m/s. If it accelerates at a rate of 5 (m/s)/s for 4 seconds, how far from the mark does it end up? 
df = di + vit + 1/2at2 
df = 10 + 3(4) + 1/2(5)(4)^2
df = 10 + 12 + 1/2 (5) (16)
df = 10 + 12 + 40
df = 62 m

Force:

Where F is force, a is acceleration, and m is mass, then:


F = ma

Work and Energy:

Work is found by


W = Fd
where W is work, F is force applied (not net force!), and d is displacement/distance.

Kinetic energy (KE) is found with this equation:
KE = 1/2 mv2
where KE is kinetic energy, m is mass and v is velocity.


The potential gravitational energy can be found with this equation:
PE = mgh
where PE is potential gravitational energy, m is mass, g is acceleration due to gravity, and h is height.

On earth, acceleration due to gravity is 9.8 (m/s)/s


The amount of elastic potential energy is determined by how hard it is to compress or stretch something and how far it is stretched or compressed.

The equation to find this is:
PE = 1/2kd2
where PE is potential elastic energy, k is a constant specific to a particular stretchy thing (spring, rubber band, etc.) and d is the distance that it is stretched or compressed (sometimes x is used instead of d, as in the illustration)



Heat/Energy Transfer

Now to find heat, we can use the formula:
Q = mcΔT
Where Q is heat or thermal energy, m is mass, c is a number called specific heat that you either look up or calculate, and ΔT is the change in temperature.



EXAMPLES:

How much heat is absorbed by 200 grams of water that starts at 25C and ends up at 30C, given that the specific heat of water is 4.2 J/g°C?

Q = mcΔT
Q = 200•4.2•(30-25)
Q = 200•4.2•5
Q = 4200 J


What is the specific heat of a metal that has a mass of 50 grams and changes temperature from 100C to 30C and gives off 4200 J of thermal energy?

Q = mcΔT
4200 = 50•c•(100-30)
4200 = 50•70•c
4200 = 3500c
4200/3500 = c
1.2 = c






VARIOUS EXAMPLES



Given the following information, find the work done on a 6.5 kg object after 4 seconds:


A = 16 N
B = 4 N
C = 14 N
D = 9 N

 STEP 1: Find Net Force by resolving the UpDown forces, resolving the LeftRight forces, and then using the Pythagorean Theorem:


F(net)2 = UpDown2 + LeftRight2

F(net)2 = (16 - 4)2 +(14 - 9)2
F(net)2 = (12)2 + (5)2
F(net)2 = 144 + 25
F(net)2 = 169
F(net) = 13 N

STEP 2: Find Acceleration where the force is the net force on the object:


F = ma

13 = 6.5a
13/6.5 = a
2 (m/s)/s  =  a

STEP 3: Find the distance through which the force acted due to the net force.


df = 1/2at2

df = 1/2 • 2 • 42
df = 16 m

STEP 4: Find the work done by the net force through the calculated distance.


W = Fd

W = 13 • 16
W = 208 J




To find the final velocity in the above:


Do Step 1 above.

Do Step 2 above.

STEP 3:


Vf = at

Vf = 2 • 4
Vf = 8 m/s


To find final kinetic energy, first find the final velocity (above), and then:


KE = 1/2mv2
KE = 1/2•6.5•82
KE = 1/2 • 6.5 • 64
KE = 208 J