Putting Scientific Notation to Work

Physics deals with some very large and very small numbers. To work with such numbers, you use scientific notation. Scientific notation is expressed as a number multiplied by a power of 10.

For example, suppose you're measuring the mass of an electron in the MKS system. You put an electron on a scale (in practice, electrons are too small to measure on a scale — you have to see how they react to the pull of magnetic or electrostatic forces to measure their mass), and you measure the following:

0.00000000000000000000000000000091 kg

What the heck is that? That's a lot of zeros, and it makes this number very unwieldy to work with. Fortunately, you know all about scientific notation, so you can convert the number into the following:

9.1 × 10-31 kg

That is, 9.1 multiplied by a power of 10, 10-31. Scientific notation works by extracting the power of 10 and putting it on the side, where it's handy. You convert a number to scientific notation by counting the number of places you have to move the decimal point to get the first digit in front of that decimal point. For example, 0.050 is 5.0 × 10-2 because you move the decimal point two places to the right to get 5.0. Similarly, 500 is 5.0 × 102 because you move the decimal point two places to the left to get 5.0.

Check out this practice question about scientific notation:

EXAMPLE

Q. What is 0.000037 in scientific notation?

A. The correct answer is 3.7 × 10-5. You have to move the decimal point five times to the right to get 3.7.

5. What is 0.0043 in scientific notation?

Solve It

6. What is 430,000.0 in scientific notation?

Solve It

7. What is 0.00000056 in scientific notation?

Solve It

8. What is 6,700.0 in scientific notation?

Solve It

Converting between Units

Physics problems frequently ask you to convert between different units of measurement. For example, you may measure the number of feet your toy car goes in three minutes and thus be able to calculate the speed of the car in feet per minute, but that's not a standard unit of measure, so you need to convert feet per minute to miles per hour, or meters per second, or whatever the physics problem asks for.

For another example, suppose you have 180 seconds — how much is that in minutes? You know that there are 60 seconds in a minute, so 180 seconds equals three minutes. Here are some common conversions between units:

[check] 1 m = 100 cm = 1,000 mm (millimeters)

[check] 1 km (kilometer) = 1,000 m

[check] 1 kg (kilogram) = 1,000 g (grams)

[check] 1 N (newton) = 105 dynes

[check] 1 J (joule) = 107 ergs

[check] 1 P (pascal) = 10 Ba

[check] 1 A (amp) = 0.1 Bi

[check] 1 T (tesla) = 104 G (gauss)

[check] 1 C (coulomb) = 2.9979 × 109 Fr

The conversion between CGS and MKS almost always involves factors of 10 only, so converting between the two is simple. But what about converting to and from the FPI and other systems of measurement? Here are some handy conversions that you can come back to as needed:

[check] Length:

• 1 m = 100 cm

• 1 km = 1,000 m

• 1 in (inch) = 2.54 cm

• 1 m = 39.37 in

• 1 mile = 5,280 ft = 1.609 km

• 1 Å (angstrom) = 10-10 m

[check] Mass:

• 1 kg = 1,000 g

• 1 slug = 14.59 kg

• 1 u (atomic mass unit) = 1.6605 × 10-27 kg

[check] Force:

• 1 lb (pound) = 4.448 N

• 1 N = 105 dynes

• 1 N = 0.2248 lb

[check] Energy:

• 1 J = 107 ergs

• 1 J = 0.7376 ft-lb

• 1 BTU (British thermal unit) = 1,055 J

• 1 kWh (kilowatt hour) = 3.600 × 106 J

• 1 eV (electron volt) = 1.602 × 10-19 J

[check] Power:

• 1 hp (horsepower) = 550 ft-lb/s

• 1 W (watt) = 0.7376 ft-lb/s

Because conversions are such an important part of physics problems, and because you have to keep track of them so carefully, there's a systematic way of handling conversions: You multiply by a conversion constant that equals 1, such that the units you don't want cancel out.

EXAMPLE

Q. A ball drops 5 meters. How many centimeters did it drop?

A. The correct answer is 500 centimeters. To perform the conversion, you do the following calculation:

5.0 meters × [100 centimeters/1 meter] = 500 centimeters

Note that 100 centimeters divided by 1 meter equals 1 because there are 100 centimeters in a meter. In the calculation, the units you don't want — meters — cancel out.

9. How many centimeters are in 2.35 meters?

Solve It

10. How many seconds are in 1.25 minutes?

Solve It

11. How many inches are in 2.0 meters?

Solve It

12. How many grams are in 3.25 kg?

Solve It

Converting through Multiple Units

Sometimes you have to make multiple conversions to get what you want. That demands multiple conversion factors. For example, if you want to convert from inches to meters, you can use the conversion that 2.54 centimeters equals 1 inch — but then you have to convert from centimeters to meters, which means using another conversion factor.

Try your hand at this example question that involves multiple conversions:

EXAMPLE

Q. Convert 10 inches into meters.

A. The correct answer is 0.254 m.

1. You know that 1 inch = 2.54 centimeters, so start with that conversion factor and convert 10 inches into centimeters:

10 in × [2.54 cm/1 in] = 25.4 cm

2. Convert 25.4 cm into meters by using a second conversion factor: 12. How many grams are in 3.25 kg?

2.54 cm × [1m/100 cm] = 0.254 m

13. Given that there are 2.54 centimeters in 1 inch, how many centimeters are there in 1 yard?

Solve It

14. How many centimeters are in a kilometer?

Solve It

15. How many inches are in an angstrom, given that 1 angstrom (Å) = 10 –8 cm?

Solve It

16. How many inches are in 3.0 meters, given that there are 2.54 cm in 1 inch?

Converting Times

Physics problems frequently ask you to convert between different units of time: seconds, minutes, hours, and even years. These times involve all kinds of calculations because measurements in physics books are usually in seconds, but can frequently be in hours.

EXAMPLE

Q. An SUV is traveling 2.78 × 10 –2 kilometers per second. What's that in kilometers per hour?

A. The correct answer is 100 km/hr.

1. You know that there are 60 minutes in an hour, so start by converting from kilometers per second to kilometers per minute:

2.78 × 10-2 [km/sec] × [60 sec/1 minute] = 1.67 km/minute

2. Because there are 60 minutes in an hour, convert this to kilometers per hour using a second conversion factor:

[1.67 km/1 minute] × [60 minutes/1 hour] = 100 km/hor

17. How many hours are in 1 week?

Solve It

18. How many hours are in 1 year?

Solve It

Counting Significant Figures

You may plug numbers into your calculator and come up with an answer like 1.532984529045, but that number isn't likely to please your instructor. Why? Because in physics problems, you use significant digits to express your answers. Significant digits, also often called significant figures, represent the accuracy with which you know your values.

For example, if you know only the values you're working with to two significant digits, your answer should be 1.5, which has two significant digits, not 1.532984529045, which has 13! Here's how it works: Suppose you're told that a skater traveled 10.0 meters in 7.0 seconds. Note the number of digits: The first value has three significant figures, the other only two. The rule is that when you multiply or divide numbers, the result has the number of significant digits that equals the smallest number of significant digits in any of the original numbers. So if you want to figure out how fast the skater was going, you divide 10.0 by 7.0, and the result should have only two significant digits — 1.4 meters per second.

On the other hand, when you're adding or subtracting numbers, the rule is that the last significant digit in the result corresponds to the last significant digit in the least accurate measurement. How does that work? Take a look at this addition example:

5.1
12
+ 7.73
------
24.83

So is the result 24.83? No, it's not. The 12 has no significant digits to the right of the decimal point, so the answer shouldn't have any either. That means you should round the value of the result up to 25.

REMEMBER

Zeros used just to fill out values down to (or up to) the decimal point aren't considered significant. For example, the number 3,600 has only two significant digits by default. That's not true if the value was actually measured to be 3,600, of course, in which case it's usually expressed as 3,600.; the final decimal indicates that all the digits are significant.

Rounding numbers in physics usually works the same way as it does in math: When you want to round to three places, for example, and the number in the fourth place is a five or greater, you add one to the third place (and ignore or replace with zeros any following digits).

EXAMPLE

Q. You're multiplying 12.01 by 9.7. What should your answer be, keeping in mind that you should express it in significant digits?

A. The correct answer is 120.

1. The calculator says the product is 116.497.

2. The number of significant digits in your result is the same as the smallest number of significant digits in any of the values being multiplied. That's two here (because of 9.7), so your answer rounds up to 120.

19. What is 19.3 multiplied by 26.12, taking into account significant digits?

Solve It

20. What is the sum of 7.9, 19, and 5.654, taking into account significant digits?

Solve It

Coming Prepared with Some Algebra

It's a fact of life: You need to be able to do algebra to handle physics problems. Take the following equation, for example, which relates the distance something has traveled (s) to its acceleration and the time it has been accelerated:

S = 1/2 at2

Now suppose that the physics problem asks you for the acceleration, not the distance. You have to rearrange things a little here to solve for the acceleration. So when you multiply both sides by 2 and divide both sides by t2, here's what you get:

[2/t2]· s = [2/t2] · [1/2] at2

Cancelling out and swapping sides, you solve for a like this:

a = 2s/t2

So that's putting a little algebra to work. All you had to do was move variables around the equation to get what you wanted. The same approach works when solving physics problems (most of the time). On the other hand, what if you had to solve the same problem for the time, t? You would do that by rearranging the variables like so:

t = [square root of (2s/a)]

The lesson in this example is that you can extract all three variables — distance, acceleration, and time — from the original equation. Should you memorize all three versions of this equation? Of course not. You can just memorize the first version and use a little algebra to get the rest.

The following practice questions call on your algebra skills:

EXAMPLE

Q. The equation for final speed, vf — where the initial speed is vo, the acceleration is a, and the time is t — is vf = vo + at. Solve for acceleration.

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

21. 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.

Solve It

22. The equation relating final speed, vf, to original speed, vo, in terms of acceleration a and distance s is v2f = v2o + 2as. Solve for s.

Solve It

23. The equation relating distance s to acceleration a, time t, and speed v is s = vot + [1/2] at2. Solve for v o.

Solve It

24. The equation for kinetic energy is KE = [1/2] mv2. Solve for v.

Solve It

Being Prepared with Trigonometry

Physics problems also require you to have some trigonometry under your belt. To see what kind of trig you need, take a look at Figure 1-1, which shows a right triangle. The long side is called the hypotenuse, and the angle between x and y is 90°.