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Test Bank For Applied Physics,12th Edition by Dale Ewen Neill Schurter P Erik Gundersen Chapter 1-24
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Test Bank For Applied Physics,12th Edition by Dale Ewen Neill Schurter P Erik Gundersen Chapter 1-24
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Test Bank ForApplied Physics, 12th Edition by Dale Ewen Neill Schurter P Erik Gundersen
Chapter 1-24
Chapter 1 The Physics Tool Kit
1) The metric prefix for 0.010 is ________.
Answer: centi
2) The metric symbol, or abbreviation, for milli is ________.
Answer: m
Write the following number(s) in scientific notation:
3) 45,000
Answer: 4.5 × 104
4) 0.0000742
Answer: 7.42 × 10-5
Write the following number in decimal form:
5) 3.28 × 10-4
Answer: 0.000328
6) 2.65 × 108
Answer: 265,000,000
Fill in each blank: (Round to three significant digits when necessary.)
7) 1.2 km = ________ m
Answer: 1200
8) 6.9 L = ________ ml
Answer: 6900
9) 750 μm = ________ mm
Answer: 0.75
10) 670 mm2 = ________ cm2
Answer: 6.7
11) 6500 cm3 = ________ m3
Answer: 0.0065
12) 15 kg = ________ lb
1
,Answer: 166
13) 96 in. = ________ yd
Answer: 2.67
14) 15 ft3 = ________ m3
Answer: 0.425
Determine the accuracy (the number of significant digits) in each measurement:
15) 20,900 m
Answer: 3
16) 0.04060 s
Answer: 4
17) 3.6 × 10-12 km
Answer: 2
Determine the precision of each measurement:
18) 23.9 m
Answer: 0.1 m
19) 14,050 ft
Answer: 10 ft
20) 5.0 × 104 s
Answer: 1000 s
21) Use the rules of measurement to add 18.5 m; 2070 cm; 95.25 cm; 0.045 m.
Answer: 40.2 m
22) Find the area of a rectangle 22.6 m long and 4.60 m wide.
Answer: 104 m2
23) Find the volume of a rectangular box 18.0 cm long, 9.0 cm wide, and 6.00 cm high.
Answer: 970 cm3
Express each of the following values using scientific notation:
24) 0.0001042 mi.
Answer: 1.042 × 10-4 mi
25) 6,800,000 in.
Answer: 6.8 × 106 in.
2
,26) 0.00481 L
Answer: 4.81 × 10-3 L
27) 0.000000015 gal
Answer: 1.5 × 10-9 gal
28) 127,500 cm
Answer: 1.275 × 105 cm
29) 0.00008207 km
Answer: 8.207 × ×10-5 km
30) 89,600,000 g
Answer: 8.96 × 107 g
31) 18,000 in.
Answer: 1.8 × 104 in.
Perform the following conversion(s):
32) 47.5 ft3 = ________ gal = ________ L
Answer: 355 gal = 1340 L
33) 20 m2 = ________ ft2 = ________ cm2
Answer: 2380 ft2 = 2,20,000 cm2
34) 923 m3 = ________ yd3 = ________ gal
Answer: 1210 yd3 = 243,000 gal
35) 55.0 gal = ________ L = ________ ft3
Answer: 208 L = 7.36 ft3
3
, Applied Physics, 12e (Ewen/Schurter/Gunderson)
Chapter 2 Problem Solving
1) What is the difference between a subscript and an exponent?
Answer: An exponent is a mathematical operation. A subscript is used to define a variable a
specific feature or component of a variable.
2) What is the difference between a formula and a working equation?
Answer: A formula is a basic equation, usually expressed in letters and numbers. A working
equation is created when the desired variable is isolated on one side of the equation.
3) What is the purpose of estimation when problem solving?
Answer: Estimating the expected answer in problem solving can serve as a check to make sure
the answer is correct.
4) Solve for m in the formula F = ma.
Answer: m = F/a
5) Solve for t in the formula s = 1/2 (vf + vi)t.
Answer: t = 2s / (vf + vi)
6) Solve for vf in the formula s = 1/2 (vf + vi)t.
Answer: vf = (2s/t) - vi
7) Solve for h in PE = mgh.
Answer: h = PE / mg
8) Given V= πr2h, if r = 5.0 cm and V = 250 cm2, what is h?
Answer: h = 3.2 cm
9) Given A = 1/2 bh, if b = 10.0 cm and h = 12.2 cm, what is A?
Answer: A = 61.0 cm2
10) A cone has a volume of 315 cm3 and a radius of 7.50 cm. What is its height?
Answer: h = 5.35 cm
11) A right triangle has a side of 82.4 mm and a side of 19.6 mm. Find the length of the
hypotenuse.
Answer: 84.7 mm
12) Given a cylinder with a radius of 14.4 cm and a height of 16.8 cm, find the lateral surface area.
Answer: 1520 cm2
4
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