U Can: Physics I For Dummies. Steven Holzner

      1. 504. The calculator says the product is 504.116. However, 19.3 has three significant digits, and 26.12 has four, so you use three significant digits in your answer. That makes the answer 504.

      2. 33 g. Here’s how you do the sum of 7.9, 19, and 5.654:

      The value 19 has no significant digits after the decimal place, so the answer shouldn’t either, making it 33 grams (32.554 grams rounded up).

      3. 28 m/s. The calculator says the quotient of 1.93 and 0.069 is 27.9710144928, and your answer should have units of meters divided by seconds, or meters per second. Because 1.93 has three significant digits and 0.069 has two (the leading zero doesn’t count), you use two significant digits in your answer. That makes the answer 28 meters per second.

      4. 6.2 m². The calculator says the product of 1.36 and 0.7130 is 0.96968, and if you add 5.2, you get 6.16968. Your answer has units of meters squared. Because 5.2 has only one significant digit after the decimal place, which is less than the three significant digits after the decimal place you keep after multiplying 1.36 and 0.7130, you use one significant digit after the decimal place in your answer. That makes the answer 6.2 meters squared.

Estimating accuracy

      Physicists don’t always rely on significant digits when recording measurements. Sometimes, you see measurements that use plus-or-minus signs to indicate possible error in measurement, as in the following:

      The

part (0.05 meters in the preceding example) is the physicist’s estimate of the possible error in the measurement, so the physicist is saying that the actual value is between

(that is, 5.41) meters and

(that is, 5.31 meters), inclusive. Note that the possible error isn’t the amount your measurement differs from the “right” answer; it’s an indication of how precisely your apparatus can measure – in other words, how reliable your results are as a measurement.

      Arming Yourself with Basic Algebra

      Physics deals with plenty of equations, and to be able to handle them, you should know how to move the variables in them around. Note that algebra doesn’t just allow you to plug in numbers and find values of different variables; it also lets you rearrange equations so you can make substitutions in other equations, and these new equations show different physics concepts. If you can follow along with the derivation of a formula in a physics book, you can get a better understanding of why the world works the way it does. That’s pretty important stuff! Time to travel back to basic algebra for a quick refresher.

      You need to be able to isolate different variables. For instance, the following equation tells you the distance, s, that an object travels if it starts from rest and accelerates at rate of a for a time, t:

      Now suppose the problem actually tells you the time the object is in motion and the distance it travels and asks you to calculate the object’s acceleration. By rearranging the equation algebraically, you can solve for the acceleration:

      In this case, you’ve multiplied both sides by 2 and divided both sides by t² to isolate the acceleration, a, on one side of the equation.

      What if you have to solve for the time, t? By moving the number and variables around, you get the following equation:

      Do you need to memorize all three of these variations on the same equation? Certainly not. You just memorize one equation that relates these three items – distance, acceleration, and time – and then rearrange the equation as needed. (If you need a review of algebra, get a copy of Algebra I For Dummies, by Mary Jane Sterling [Wiley].)

       Example

      Q. The equation for final speed, v_f – where the initial speed is v_i, the acceleration is a, and the time is t – is v_f = v_i + at. Solve for acceleration.

      A. The correct answer is a = (v_f – v_i)/t. To solve for a, divide both sides of the equation by time, t.

       Practice Questions

      1. The equation for potential energy, PE, for a mass m at height h, where the acceleration due to gravity is g, is PE = mgh. Solve for h.

      2. The equation relating final speed, v_f , to initial speed, v_i , in terms of acceleration a and distance s is v_f ² = v_i ² + 2as. Solve for s.

      3. The equation relating distance s to acceleration a, time t, and speed v is

. Solve for v_i .

      4. The equation for kinetic energy is

. Solve for v.

       Practice Answers

      1. h = PE/mg. Divide both sides by mg to get your answer.

      2.

. Divide both sides by 2a to get your answer.

      3.

. Subtract

from both sides:

      Divide both sides by t to get your answer.

      4.

. Multiply both sides by 2/m:

      Take the square root to get your answer.

      Tackling a Little Trig

You need to know a little trigonometry, including the sine, cosine, and tangent functions, for physics problems. To find these values, start with a simple right triangle. Take a look at Figure 2-1, which displays a right triangle in all its glory, complete with labels I’ve provided for the sake of explanation. Note in particular the angle

, which appears between one of the triangle’s legs and the hypotenuse (the longest side, which is opposite the right angle). The side y is opposite

, and the side x is adjacent to

.

      © John Wiley & Sons, Inc.

       Figure 2-1: A labeled triangle that you can use to find trig values.

       Remember: To find the trigonometric values of the triangle in Figure 2-1, you divide one side by another. Here are the definitions of sine, cosine, and tangent:

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