A01_Math Review 1

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Apr 3, 2024

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Old Dominion University Physics 103N Laboratory Manual 1 Math Review Experiment A01 Kaleb Name Lab Section Objective Review the following selected mathematical topics: Importance of Units SI Units & the Metric System Unit Conversion Scientific Notation Order of Operations Percentage Difference Materials Calculator Theory See the appendix to your lab manual for the theory section to this experiment. It is available for download on Canvas. Order of Operations 1. Solve the following equations. If needed, write each answer to 3 decimal places. Operation Value a. 2 + 3 – 5 ÷ 8 4.375 b. (10 ÷ 2) 2 + 12 - 3 3 10 c. (12 ÷ 3) 3 + 10 - 4 x 6 50 d. (6 2 + (8 ÷ 4 + 3 2 )) -4 2 31 Checking Your Calculator 2. One equation you will see a lot in your astronomy class is related to Kepler’s Third Law of Planetary Motion. It relates how long it takes a planet to order the Sun to how far away it orbits. Simply, it is This is due for everyone! Not as a group.
2 Experiment A01: Math Review P 2 = k a 3 ( for a ) Let’s use Jupiter as an example. It takes 11.862 Earth years for Jupiter to orbit the Sun once (P = 11.862 years), and its orbits about 5.2 times further out than the Earth (a = 5.2) If we solve the previous equation for k , we find k = P 2 a 3 Use this equation to determine the value of k. Give your answer to 4 decimal places. (Hint: Your answer will be very close to 1. This is just a check that you can properly put in the steps in your calculator. If you can’t get an answer close to 1, then you should check how you type your this into your calculator.) 140.701/140.608=1.0007 Scientific Notation 3. Practice with scientific notation – Write out the decimal equivalent (regular form) of the following numbers that are in scientific notation and vice versa. 4. Perform the indicated operation and express the answer in scientific notation. Operation Scientific Notation a. (0.314) × (0.006313× 10 6 ) 1.982282 x 10^3 b. (1.11265 × 10 -13 ) + (2.22 × 10 -16 ) 1.11487 x 10^-13 c. (3.3456 × 10 20 ) ÷ (2.244 × 10 25 ) 1.490 x 10^-5 d. (1.999 × 10 6 ) × (1.999 × 10 9 ) 3.996001 x 10^15 Scientific Notation Decimal Equivalent Scientific Notation Decimal Equivalent 10 2 100 3 x 10 2 300 10 4 10000 7 x 10 4 70000 10 7 10,000,000 2.4 x 10 3 2400 10 -2 0.01 6 x 10 -3 0.006 10 -5 0.00001 900 x 10 -2 9 10 0 1 4 x 10 -6 0.000004 Scientific Notation Decimal Equivalent Scientific Notation Decimal Equivalent 10 3 1000 2 x 10 3 2,000 a. 10^1 10 a. 4 x 10^2 400 b. 10^2 100 b. 6 x10^4 60,000 c. 10^8 100,000,000 c. 7.5 x 10^5 750,000 d. 10^-1 0.1 d. 5 x 10^3 0.005 e. 10^-4 0.0001 e. 3.4 x 10^-3 0.0034 f. 10^0 1 f. 6.457 x 10^- 2 0.06457
Old Dominion University Physics 103N Laboratory Manual 3 Unit Conversion 5. Convert the following values to other appropriate units. Value Converted Value Converted Value a. 12 inches 1 feet 0.000189394 miles b. 4 feet 48 inches 1.2192 meters c. 27 inches 68.85 cm 0.6858meters d. 1 yard 36 inches 0.9144 meters e. 1 meter 100 centimeters 1000 millimeters For analysis questions 6 through 8, express your answer in scientific notation. 6. There are an estimated 10 11 stars in a typical galaxy, and there are an estimated 10 11 galaxies in the Universe. How many stars are there in the Universe? Show your work. 10^22 7. If there are 2.8 × 10 8 people in the US, and there are 4 × 10 7 people in Canada, how many people are there in the US and Canada combined? How many more people live in the US than in Canada? Show your work. 2.8 x 10^8 + 4 x 10^7 = 320,000,000 people (3.2 x 10^8) 2.8 x 10^8 = 280000000 4 x 10^7 = 40000000 280000000 - 40000000= 240000000 (2.4 x 10^8)
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4 Experiment A01: Math Review 8. There are 6.02 × 10 23 atoms in the chemical quantity called a mole. How many atoms are in ½ of a mole? How many atoms in one hundredth of a mole? Show your work. 0.5 (6.02 x 10^23) = (Atoms that are ½ of a mole) 3.01 x 10^23 (6.02 x 10^23)0.01 = 6.02 x 10^21 (Atoms that are in one hundredth of a mole) Algebraic Solutions 9. Solve the following equations for the variable indicated. a) v = x t ( for t ) b) F = G mM R 2 ( for R ) c) L = σAT 4 ( for T ) a). T f x/v. b. sqrt(gmMf/f) Percent Error 10. Calculate the percent error (% error) for the following case: During their experiment, two lab partners conducted several trials to determine the value of acceleration of a racquet ball dropped from a height of 2 meters and they found that the average value of acceleration was 9.78 m/s 2 . However, their reference manual states that the accepted value of free-fall acceleration is 9.81 m/s 2 . Find the percentage Error between the accepted value and the measured value. Show your work. The percent error between the accepted value and the measured value is approximately 0.3056% 0.03/9.81 x 100 = approximately 0.3056%
Old Dominion University Physics 103N Laboratory Manual 5