Physics Phenomena "Physics is Fun" Feimer's Physics Page Links To Student Infomation Class and Club Information: Astronomy Class and Club Information: Physics Class & Study Schedule Information: Return to Top of Page Physics Math Review Mathematics is the foundation of physics.  The level of mathematics necessary to comprehend the technical, problem solving aspect of physics varies and is dependent upon the level at which a "student" is pursuing the subject.  Concepts in physics can be studied and investigated in a non technical, non mathematical approach, where, at most, arithmetic is used.  This is the level at which a child or adult who has a curiosity about the science, but no background knowledge would be expected to start. At the high school level a basic knowledge of algebra is expected for the student in a typical first course.  The small amount of geometry and trigonometry that might be necessary in a high school course can be taught in the course as they are needed.  A more advanced high school course such as what is often called an honors level physics course would usually require algebra, geometry, and either concurrent enrollment in or successful completion of trigonometry as prerequisites for the course. High School Physics courses are generally divided into two categories. These are the advanced placement, Calculus based physics course, and the Algebra based physics course. The student who is planning to major in physics or engineering will want to enroll in the calculus based physics course, other students who want to learn about physics, but do not intend to major in phyiscs related subjects will find the algebra based physics to be the route to go. Math Review Topics: The following is a list of six (6) categories of basic math skills (with instructional information) recommended for anyone interested in studying physics. The topics from which you may select from are:

 1. Solving for a single variable in a first degree equation. In an equation such as  a = b / c  being able to solve for a, b, or c correctly is extremely important.  The most common mistake in this situation is thinking that the solution in the case of solving for c is not recognizing that  c = b / a .  Way too often students think that the value for c is equal to a / b.  The most basic rule for solving algebraic equations of any size or complexity is that: "You must always do the same thing to both sides of an equation, regardless of what operation you are applying." When a student thinks that c = a / b in the above equation rather than c = b / a, it is because they did NOT apply this rule. To begin consider the ratio and proportion equation a / b = c / d.  Step 1. When solving for any variable in this type of equation first cross multiply. This means that a is multiplied times d, and b is multiplied times c. This yields a d = b c Step 2. To solve for any one of the four variables, the variable being solved for must be isolated by the use of division. For Example, if the variable being solved for was d, both sides of the equation would be divided by a. This yields d = b c / a   As a similar example, consider the following. The equation  a = b / c  can be solved for c by the following two steps.  Step 1.  Multiply both sides of the equation by c This yields  c a = c b / c This simplifies to  c a = b Step 2.  Divide both sides of this equation by c This yields  c a / c  = b / c This simplifies to  a = b / c Thus solving for c  in the expression (equation)  a = b / c results in the expression (equation)  c = b / a [Note: For convenience a could be written over 1, as in a / 1 = b / c, and c could be solved for using a ratio and proportion approach.]   Additional examples of solving for variables: As stated above the most important rule to focus on in solving any equation is to avoid destroying an equality by doing some operation to one side that you do not do to the other side. To see the significance of this statement look at the following examples. Example 1: solve for "v" in the equation v = d / t  The "d" is a numerator (a value in the top of a fraction)  The "t" is a denominator (a value in the bottom of a fraction)  The "v" is equal to "d" divided by "t" Example 2: solve for "d" in the equation v = d / t  The "d" is a numerator  The "t" is a denominator  "v" is equal to "d" divided by "t"  "d" can be solved for by multiplying both sides by "t"  This gives us v t = d or d = v t Example 3: solve for t in the equation v = d / t  The "d" is a numerator  The "t" is a denominator  The "v" is equal to "d" divided by "t"  "t" can be solved for by   1st multiplying both sides by "t" which gives    v t = d   2nd dividing both sides by "v" which gives    t = d / v The second rule is to be patient when solving equations which will require that you use several steps. Example 4: Solve for "t" in the equation v(f) = v(i) + at  There are no fractions here but there is an addition  of unlike variables on the right side of the equation.  "t" can be solved for by   1st subtracting "v(i)" from both sides which gives    v(f) - v(i) = at   2nd divide both sides by "a" which gives    a = v(f) - v(i) / a = t Return to Top of Page

 3. Measurement and the Metric System The metric system was developed around 1775 by French scientists. It is convenient to use because its units are related by powers of ten. A standardized system now exists worldwide. It is referred to as the International System (SI) of weights and measures. Traditionally in the study of physics, two systems of measure have been used. They are referred to as the MKS and the CGS systems. MKS system meter - kilogram - second CGS system centimeter - gram - second Many textbooks emphasize the MKS system The Metric System: Four basic units of measure are displayed. Length:  base unit = meter (fundamental unit)* Mass:  base unit = gram (fundamental unit)* Time:  base unit = second (fundamental unit)* Volume:  base unit = liter (derived unit)* * Units of measure are divided into two categories called the Fundamental unit and the Derived Unit. Fundamental Unit: These are units of measure obtained directly from measurement observations using SI units and cannot be obtained indirectly from any simpler or more basic units of measure. Mass, length, and time are examples of fundamental units. Derived Unit: These are units of measure which may be able to be observed through direct measurement (i.e. volume can be observed directly by visibly observing the reading off of the side of a graduate cylinder.), but also can be obtained by derivation using simpler fundamental units of measure. As an example consider that volume can be obtained from length by the formula V = L x W x H. Example: Length of course is a length measurement (L), but so are width (W) and height (H) length measurements simply oriented in different directions. And it follows then that volume, which is a multiplication product of the three, is derived from length. - - - - - - - - - - - - - Metric Prefixes - Units of Measure Prefix Symbol Fractional Equivalent Example (using the meter) Kilo K 1000 x base unit kilometer (Km) hecto h 100 x base unit  hectometer (hm) deka da 10 x base unit  dekameter (dam) ___ ___ 1 x base unit  meter (m) deci d 1/10 x base unit decimeter (dm) centi c 1/100 x base unit centimeter (cm) milli m 1/1000 x base unit millimeter (mm) - - - - - - - - - - - - - Additional Units: Prefix (Symbol) Fractional Equivalent Example (using the meter) giga (G) 1 x 10^9 x base unit gigameter (Gm) mega (M) 1 x 10^6 x base unit megameter (Mm) micro (m *) 1/1 x 10^-6 x base unit micrometer (mm *) nano (n) 1/1 x 10^-9 x base unit nanometer (nm) pico (p) 1/1 x 10^-12 x base unit picometer (pm) * This is the Greek letter mu and not the English letter u. - - - - - - - - - - - - - Common Metric - English Equivalents: 1 m = 39.37 in 1 cm = 0.394 in 1 in = 2.54 cm 1 kg = 2.21 lb 1 lb = 454 g 1 L = 1.057 qt 1 oz = 28.4 g 1 in^3 = 16.5 cm^3 1 gal = 3.78 L 1-8 Metric Conversion - Metric Sliding Scale Model - - - - - - - - - - - - - Conversions between metric units: The metric system is very easily manipulated as can be seen by the following two examples. Converting between two units of the same category of measurement can be done simply by moving the position of the decimal place at the same time the prefix is changed.  As the unit of measure is increased the decimal is moved to the left making the numerical value smaller proportionally. As the unit of measure is decreased the decimal is moved to the right making the numerical value larger proportionally. Look at the following two examples of this concept. Example 1:  Consider the measurement 2.5 m and the variations of the same length measurement expressed in other metric units of length. km                 hm              dam          b. u.*       dm            cm             mm 0.0025          0.025          0.25          2.5          25.0          250.          2,500. *b. u. is the base unit of the category of measurement.  In this case the unit is the meter. Example 2:  Consider the measurement 4,200 m and the variations of the same length measurement expressed in other metric units of length. km          hm           dam           b. u.            dm                 cm                    mm 4.2          42.          420.          4,200.          42,000.          420,000.          4,200,000. Return to Top of Page

 4. Working with multiple dimensional conversions. Student who study doing conversions between units within a measurement system such as Meteric to Metric or between measuring systems such as the Metric to English and English to Metric conversions usually can handle conversion factors that are one dimensional.  However, a problem often arises when conversion factors are needed for two and three dimensional problems.  The following information compares handling units in one, two, and three dimensions where conversions are involved. - - - - - - - - - - - - - 1. One Dimensional conversion. Given a length of 100 inches, convert this to centimeters. The one dimensionsl (linear) conversion factor is 1 in = 2.54 cm. The calculation involves  100 in x 2.54 cm / in = 254 cm. - - - - - - - - - - - - - 2. Two Dimensional conversion. Given a surface area of 500 square inches (500 in^2), convert this to square centimeters. The two dimensional (Area) conversion factor is (1 in)^2 = (2.54 cm)^2  or 1 in^2 = 6.4516 cm^2 The calculation involves  500 in^2 x 6.4516 cm^2 / in^2 = 3,225.8 cm^2 - - - - - - - - - - - - - 3. Three Dimensional conversion. Given a volume of 1200 cubic inches (1200 in^3), convert this to cubic centimeters. The three dimesnional (Volume) conversion factor is (1 in)^3 = (2.54 cm)^3  or  1 in^3 = 16.387064 cm^3 The calculation involves  1200 in^3 x 16.387064 cm^2 / in^3 = 19,664.4768 cm^3 Return to Top of Page

 5. Dimensional Analysis, also called Factor Labeling. Dimensional analysis focuses on the units of measurement, which is an area of mathematics that many students overlook.  Dimensional analysis focuses on the use of measurements in calculations.  Below are some examples of using units in calculations. - - - - - - - - - - - - - 1. Calculating the surface area of a shelf board. Given a 24 inch by 48 inch shelf board calculate the amount of surface area available to store objects on. Equation to be used:   Area = length x width Substitution:          Area = 48 inches x 24 inches Answer:                Area = 1,152 square inches  or  1,152 in^2 - - - - - - - - - - - - - 2. Converting the surface area determined in the above problem into square feet. The shelf board above has a surface area of  1,152 in^2 There are two ways to go about this.  One is to convert the original dimensions to feet and then substitute into the equation and solve.  The other way is to solve for the answer by doing a conversion calculation. The 1st way:  The length is  48 in x 1 ft / 12 in = 4 ft                        The width is  24 in x 1 ft / 12 in = 2 ft Calculation of Area:     A = L x w = 4 ft x 2 ft = 8 square feet  or  8 ft^2 The 2nd way:  The area is  1,152 in^2 The conversion factor between the inch and the foot is  12 in = 1 ft The conversion factor between the square inch and the square foot involves squaring both sides of the above conversion factor.  This gives us: Conversion factor in^2 --> to ft^2 is  (12 in)^2 = (1 ft)^2 Using this conversion factor:  1,152 in^2 x (1 ft^2 / 144 in^2) Answer:     8 ft^2 - - - - - - - - - - - - - 3. The same approach is used in volume problems. Calculate the volume of a box having the following dimensions.      48 inches long, 24 inches wide, 36 inches deep (or high) Equation to be used:     Volume = length x width x depth (or height) Substitution:            Volume = 48 in  x  24 in  x  36 in Answer:                  Volume = 41,472 in^3 - - - - - - - - - - - - - 4. Determining the volume in cubic feet would involve the following process. Again we could approach this in two different ways. The 1st way:     The length is 48 in x 1 ft / 12 in = 4 ft                          The width is 24 in x 1 ft / 12 in = 2 ft                          The depth is 36 in x 1 ft / 12 in = 3 ft (also height) Calculation of Volume:     V = L x W x H = 4 ft x 2 ft x 3 ft = 24 cubic feet  or  24 ft^3 The 2nd way:     The area is  41,472 in^3 The conversion factor between the inch and the foot is  12 in = 1 ft The conversion factor between the cubic inch and the cubic foot involves cubing both sides of the above conversion factor.  This gives us: Conversion factor in^3 --> to ft^3 is  (12 in)^3 = (1 ft)^3 Using this conversion factor:  41,472 in^2 x (1 ft^2 / 1,728 in^2) Answer:     24 ft^3 - - - - - - - - - - - - - 5. Determining the Density of a substance. The mass of a rectangular solid piece of metal is  100 g Its dimensions are:  Length is 5 cm, Width is 4 cm, and thickness (height) is 2 cm. Equation used:     Density = Mass / Volume Volume is:     V = L x W x H = 5 cm x 4 cm x 2 cm = 40 cm^3 Substitution:     D = M / V = 100 g / 40 cm^3 = 2.5 g/cm^3 Note:  Density is a derived unit with a complex unit of measure. Return to Top of Page