1.1 Starting with the definition find the number of (a) kilometers in 1.00 mile and (b) feet in 1.00 km.
Answer:
1.1. IDENTIFY: Convert units from mi to km and from km to ft.
SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.
EXECUTE:
(a) 1.00 mi = (1 00 mi) ((5280 ft)/(1 mi))((12 in)/(1 ft))((2.54 cm)/(1 in))(1m/(〖10〗^2 cm))(1km/(〖10〗^3 m)) = 1.61 km
SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.
EXECUTE:
(a) 1.00 mi = (1 00 mi) ((5280 ft)/(1 mi))((12 in)/(1 ft))((2.54 cm)/(1 in))(1m/(〖10〗^2 cm))(1km/(〖10〗^3 m)) = 1.61 km
(b) 1.00 km = (1.00 km) ((〖10〗^3 m)/1km)((〖10〗^2 cm)/1m)((1 in)/(2.54 cm))((1 ft)/(12 in)) = 3.28 × 103 ft
EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.
1.2 According to the label on a bottle of salad dressing, the volume of the contents is 0.473 liter (L). Using only the conversions and express this volume in cubic inches.
EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.
1.2 According to the label on a bottle of salad dressing, the volume of the contents is 0.473 liter (L). Using only the conversions and express this volume in cubic inches.
Answer:
1.2. IDENTIFY: Convert volume units from L to in.3.
SET UP: 1 L =1000 cm3. 1 in. = 2.54 cm
EXECUTE:
0 473 L×((〖10〗^3 cm)/1L)×((1 in.)/(2.54 cm))^3 = 28.9 in.3
EVALUATE: 1 in.3 is greater than 1 cm3, so the volume in in.3 is a smaller number than the volume in cm3, which is 473 cm3.
SET UP: 1 L =1000 cm3. 1 in. = 2.54 cm
EXECUTE:
0 473 L×((〖10〗^3 cm)/1L)×((1 in.)/(2.54 cm))^3 = 28.9 in.3
EVALUATE: 1 in.3 is greater than 1 cm3, so the volume in in.3 is a smaller number than the volume in cm3, which is 473 cm3.
1.3 How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)
Answer:
1.3. IDENTIFY: We know the speed of light in m/s. t = d/v. Convert 1.00 ft to m and t from s to ns.
SET UP: The speed of light is v = 3.00 ×10^8 m/s. 1 ft = 0.3048 m. 1 s =10^9 ns.
EXECUTE:
t = (0.3048 m)/(3.00×〖10〗^8 m/s) = 1.02×〖10〗^(-9) = 1.02 ns
EVALUATE: In 1.00 s light travels 3.00×〖10〗^8 m = 3.00×〖10〗^5 km =1.86×〖10〗^5 mi.
1.4 The density of gold is What is this value in kilograms per cubic meter?
SET UP: The speed of light is v = 3.00 ×10^8 m/s. 1 ft = 0.3048 m. 1 s =10^9 ns.
EXECUTE:
t = (0.3048 m)/(3.00×〖10〗^8 m/s) = 1.02×〖10〗^(-9) = 1.02 ns
EVALUATE: In 1.00 s light travels 3.00×〖10〗^8 m = 3.00×〖10〗^5 km =1.86×〖10〗^5 mi.
1.4 The density of gold is What is this value in kilograms per cubic meter?
Answer:
1.4. IDENTIFY: Convert the units from g to kg and from cm3 to m3.
SET UP: 1 kg =1000 g. 1 m =1000 cm.
EXECUTE:
19.3 g/cm^3 ×((1 kg)/1000g)×(100cm/1m)^3 = 1.93×〖10〗^4 kg/m^3
SET UP: 1 kg =1000 g. 1 m =1000 cm.
EXECUTE:
19.3 g/cm^3 ×((1 kg)/1000g)×(100cm/1m)^3 = 1.93×〖10〗^4 kg/m^3
EVALUATE: The ratio that converts cm to m is cubed, because we need to convert 〖cm〗^3 to m^3.
1.5 The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters (L) by using only the conversions and 1 in. = 2.54 cm.
Answer:
1.5. IDENTIFY: Convert volume units from in.3 to L.
SET UP: 1 L =1000 cm3. 1 in. = 2.54 cm.
EXECUTE: (327 in.3) × (2.54 cm/in.)3 × (1 L/1000 cm3) = 5.36 L
EVALUATE: The volume is 5360 cm3. 1 cm3 is less than 1 in.3, so the volume in cm3 is a larger number than the volume in in.3.
1.6 A square field measuring 100.0 m by 100.0 m has an area of 1.00 hectare. An acre has an area of If a country lot has an area of 12.0 acres, what is the area in hectares?
Answer:
IDENTIFY: Convert ft2 to m2 and then to hectares.
SET UP: 1.00 hectare =1.00×104 m2. 1 ft = 0.3048 m.
EXECUTE: The area is (12 0 acres)(〖43 600 ft 〗^2/(1 acres)) ((0.3048 m)/(1.00 ft))^2 ((1.00 hectare)/(1.00×〖10〗^4 m^2 ))= 4.86 hectares.
EVALUATE: Since 1 ft = 0.3048 m, 1 ft2 = (0.3048)2 m2.
1.7 How many years older will you be 1.00 gigasecond from now? (Assume a 365-day year.)
Answer:
IDENTIFY: Convert seconds to years.
SET UP: 1 billion seconds =1×109 s. 1 day = 24 h. 1 h = 3600 s.
EXECUTE: 1.00 billion seconds = (1.00 109 s) (1h/3600s)(1d/(24 h))(1y/(365 d)) = 31.7 y.
EVALUATE: The conversion 1 y = 3.156 ×107 s assumes 1 y = 365.24 d,which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year.
1.8 While driving in an exotic foreign land you see a speed limit sign on a highway that reads 180,000 furlongs per fortnight. How many miles per hour is this? (One furlong is and a fortnight is 14 days. A furlong originally referred to the length of a plowed furrow.)
Answer:
IDENTIFY: Apply the given conversion factors.
SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.
EXECUTE: (180 000 furlongs/fortnight) ((0.125 mi)/(1 furlong ))((1 fortnight)/(14 d))((1 d)/(24 h)) = 67 mi/h
EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number.
1.9 A certain fuel-efficient hybrid car gets gasoline mileage of 55.0 mpg (miles per gallon). (a) If you are driving this car in Europe and want to compare its mileage with that of other European cars, express this mileage in Use the conversion factors in Appendix E. (b) If this car’s gas tank holds 45 L, how many tanks of gas will you use to drive 1500 km?
Answer:
IDENTIFY: Convert miles/gallon to km/L.
SET UP: 1 mi =1.609 km. 1 gallon = 3.788 L.
EXECUTE: (a) 55.0 miles/gallon = (55.0 miles/gallon) ((1.609 km)/1mi)((1 gallon)/(3.788 L)) = 23.4 km/L.
(b) The volume of gas required is (1500 km)/(23 4 km/L) = 64 1 L.
(64.1 L)/(45 L/tank) = 1.4 tanks.
EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in a
gallon, so 1 mi/gal ~2/4 km/L, which is roughly our result.
1.10 The following conversions occur frequently in physics and are very useful. (a) Use 1 mi = 5280 ft and 1 h = 3600 s to convert 60 mph to units of ft/s (b) The acceleration of a freely falling object is 32ft/s2. Use 1 ft = 30.48 cm to express this acceleration in units of m/s2. (c) The density of water is 1.0 g/cm3. Convert this density to units of kg/m3.
IDENTIFY: Convert units.
SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g =1 kg.
EXECUTE: (a) ((60 mi)/h)((1 h)/3600s)((5280 ft)/1mi)=88 ft/s
1.5 The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters (L) by using only the conversions and 1 in. = 2.54 cm.
Answer:
1.5. IDENTIFY: Convert volume units from in.3 to L.
SET UP: 1 L =1000 cm3. 1 in. = 2.54 cm.
EXECUTE: (327 in.3) × (2.54 cm/in.)3 × (1 L/1000 cm3) = 5.36 L
EVALUATE: The volume is 5360 cm3. 1 cm3 is less than 1 in.3, so the volume in cm3 is a larger number than the volume in in.3.
1.6 A square field measuring 100.0 m by 100.0 m has an area of 1.00 hectare. An acre has an area of If a country lot has an area of 12.0 acres, what is the area in hectares?
Answer:
IDENTIFY: Convert ft2 to m2 and then to hectares.
SET UP: 1.00 hectare =1.00×104 m2. 1 ft = 0.3048 m.
EXECUTE: The area is (12 0 acres)(〖43 600 ft 〗^2/(1 acres)) ((0.3048 m)/(1.00 ft))^2 ((1.00 hectare)/(1.00×〖10〗^4 m^2 ))= 4.86 hectares.
EVALUATE: Since 1 ft = 0.3048 m, 1 ft2 = (0.3048)2 m2.
1.7 How many years older will you be 1.00 gigasecond from now? (Assume a 365-day year.)
Answer:
IDENTIFY: Convert seconds to years.
SET UP: 1 billion seconds =1×109 s. 1 day = 24 h. 1 h = 3600 s.
EXECUTE: 1.00 billion seconds = (1.00 109 s) (1h/3600s)(1d/(24 h))(1y/(365 d)) = 31.7 y.
EVALUATE: The conversion 1 y = 3.156 ×107 s assumes 1 y = 365.24 d,which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year.
1.8 While driving in an exotic foreign land you see a speed limit sign on a highway that reads 180,000 furlongs per fortnight. How many miles per hour is this? (One furlong is and a fortnight is 14 days. A furlong originally referred to the length of a plowed furrow.)
Answer:
IDENTIFY: Apply the given conversion factors.
SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.
EXECUTE: (180 000 furlongs/fortnight) ((0.125 mi)/(1 furlong ))((1 fortnight)/(14 d))((1 d)/(24 h)) = 67 mi/h
EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number.
1.9 A certain fuel-efficient hybrid car gets gasoline mileage of 55.0 mpg (miles per gallon). (a) If you are driving this car in Europe and want to compare its mileage with that of other European cars, express this mileage in Use the conversion factors in Appendix E. (b) If this car’s gas tank holds 45 L, how many tanks of gas will you use to drive 1500 km?
Answer:
IDENTIFY: Convert miles/gallon to km/L.
SET UP: 1 mi =1.609 km. 1 gallon = 3.788 L.
EXECUTE: (a) 55.0 miles/gallon = (55.0 miles/gallon) ((1.609 km)/1mi)((1 gallon)/(3.788 L)) = 23.4 km/L.
(b) The volume of gas required is (1500 km)/(23 4 km/L) = 64 1 L.
(64.1 L)/(45 L/tank) = 1.4 tanks.
EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in a
gallon, so 1 mi/gal ~2/4 km/L, which is roughly our result.
1.10 The following conversions occur frequently in physics and are very useful. (a) Use 1 mi = 5280 ft and 1 h = 3600 s to convert 60 mph to units of ft/s (b) The acceleration of a freely falling object is 32ft/s2. Use 1 ft = 30.48 cm to express this acceleration in units of m/s2. (c) The density of water is 1.0 g/cm3. Convert this density to units of kg/m3.
IDENTIFY: Convert units.
SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g =1 kg.
EXECUTE: (a) ((60 mi)/h)((1 h)/3600s)((5280 ft)/1mi)=88 ft/s
(b) ((32 ft)/s^2 )((30.48 cm)/1ft)(1m/100cm)=9.8 m/s^2
(c) ((1.0 g)/〖cm〗^3 ) (100cm/1m)^3 (1kg/1000g)=〖10〗^3 kg/m^3
EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 =103 kg/m3 are exact. The relation
32 ft/s2 = 9.8 m/s2 is accurate to only two significant figures.
(c) ((1.0 g)/〖cm〗^3 ) (100cm/1m)^3 (1kg/1000g)=〖10〗^3 kg/m^3
EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 =103 kg/m3 are exact. The relation
32 ft/s2 = 9.8 m/s2 is accurate to only two significant figures.
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