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cricket20 [7]
3 years ago
15

if a tourist walked all 3 trails in a day,how many kilometers would he or she have walked? km, (2.39, 1.28, 2.542)

Mathematics
1 answer:
natita [175]3 years ago
4 0
He or she would have walked a total of all three numbers for example 2.39 + 1.28 would be 3.67, add 2.542 to that to get the total of 6.202 km that he/she walked.


   hope this helps  :)
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CONVERSIONS, PHYSICAL CONSTANTS, AND UNITS

Length:
1 inch (in) =  2.54 centimeters (cm) exactly
1 foot (ft) =  0.305 meters (m)
1 yard (yd)  = 0.914 meters (m)
1 mile (mi) = 5280 feet (ft)  =  1.61 kilometers (km)
1 centimeter (cm)  = 0.394 inches (in)
1 meter (m)  = 100 centimeters (cm)  =  39.4 inches (in) = 3.28 feet (ft)  =  1.09 yards (yd)
1 kilometer (km) = 1000 meters (m)  = 0.621 miles (mi)
1 rod =  16.5 feet (ft) = 5.5 yards (yd) =  5.03 meters (m)
1 astronomical unit (AU) = 1.50 x 108 kilometers (km)  = 9.29 x 107 miles (mi)  
1 light year (LY) = 9.461 x 1012 kilometers ( km) = 5.88 x 1012 miles (mi)  = 0.307 parsec
1 parsec =  approximately 3.26 light years (LY)

Area  =  length  x  width
1 square inch (in2) =  0.00694 square feet (ft2  =  6.45 square centimeters, cm2
1 square foot, (ft2) =  144 square inches (in2  =  0.0929 square meters, m2
1 square yard (yd2  =  1296 square inches (in2  =  9 square feet (ft2  =  0.836 square meters (m2
1 square mile (mi2  =  640 acres  =  2.59 square kilometers (km2 
1 acre  = 160 square rods  =  43,560 square feet (ft2) =  0.405 hectares
1 hectare  =  10,000 square meters (m2  =  100 ares  =  2.47 acres
1 square rod  =  30,25 square yards (yd2  =  25.29 square meters (m2
1 are  =  100 square meters (m2  =  0.01 hetares  =  119.6 square yards (yd2   

Volume =  length  x  width  x  height
1 cubic inch (in3) =  0.000579 cubic feet (ft3) =  16.4 cubic centimeters (cm3)
1 cubic foot (ft3) =  1728 cubic inch (in3) =  0.0283 cubic meters (m3)
1 cubic yard (yd3) =  27 cubic feet (ft3) =  4.65 x 104 cubic inch (in3) =  0.765 cubic meters (m3)
1 cubic meters (m3) =  106 cubic centimeters (cm3) =  1,000 liters (L)  =  35.3 cubic feet (ft3)
1 quart (qt)  =  2 pints (pt) = 946 milliliters (mL) = 0.946 liters (L)
1 gallon (gal) = 4 quarts  = 231 cubic inch (in3) =  3.79 liters (L)
1 liter (L) =  1000 cubic centimeters (cm3) = 1.06 quarts (qt) = 0.265 gallons (gal)

Mass
1 slug = 14.6 kilograms (kg)
1 kilogram (kg) =  1,000 grams  = 0.0685 slugs 
1 atomic mass unit (u)  =  1.66 x 10-27 kilogram (kg) =1.66 x 10-24
1 electron mass  = 9.11 x 10-31 kilogram (kg) =  9.11 x 10-28 grams (g)  =  5.46 x 10-4 atomic mass unit (u) 
1 proton mass =  1.00728 atomic mass unit (u)
1 neutron mass =  1.00866 atomic mass unit (u)

Conversion Between Weight and Mass on Earth
A slug is the mass unit in the British system and is equal to 32.2 pounds (lb) 
A kilogram weighs 9.81 newtons (N) or 2.21 pounds (lb) 
A mass of one (1) gram (g) weighs 981 dynes or 0.0353 ounces (oz) 
A one (1) ounce weight has a mass 28.4 grams (g) or 0.0284 kilograms (kg) 
A one (1) pound weight has a mass of 454 grams (g) or 0.454 kilograms (kg) 
A metric ton has a mass of 1,000 kilograms (kg) 
A one (1) ton (regular) weight has a mass of 2,000 pounds (lb)

Velocity and Speed  =  distance/time
1 ft/s (ft s-1)  =  0.305 m/s (m s-1)
1 mi/hr (mi hr-1) =  1.47 ft/s (ft s-1) = 1.61 km/hr (km hr-1)  = 0.447 m/s (m s-1)
1 m/s (m s-1)  =  100 cm/s (cm s-1) =  3.28 ft/s (ft s-1)
1 km/hr (km hr-1) =  0.278 m/s (m s-1) = 0.621 mi/hr (mi hr-1)  = 0.912 ft/s (ft s-1)
Speed of light in a vacuum  = 3.00 x 108 m/s (m s-1) =  3.00 x 1010 cm/s (cm s-1)  = 186,000 mi/s (mi s-1)
Speed of sound in air at 0 oC (32 oF) =  331 m/s  (m s-1) =  1090 ft/s (ft s-1)  = 41 mi/hr (mi hr-1)

Acceleration =  velocity /  time
1 ft/s2  = 0.305 m/s2  =  30.5 cm/s2
1 mi/hr/s  = 1.47 ft/s2 =  1.61 km/hr/s  =  0.447 m/s2
1 m/s2 = 100 cm/s2 = 3.28 ft/s2
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Force  = mass x acceleration
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1 N (kg· m/s2)  =  0.223 lb

Temperature
oF =  9/5  x  oC  + 32
oC = (oF  -  32)  x  5/9
Absolute Temperature, kelvin, K  =  oC  +  273.15
0 K  =  -273.15 oC  =  -459.72 oF
Normal Body Temperature  =  98.6 oF

Pressure  =  Force/Area
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1 bar  =  105 N/m2  =  14.50 lb/in2
1 atmosphere, atm  =  760 mm Hg  =  76.0 cm Hg  =  760 torr  =  14.50 lb/in2  =  1.01325 x 105 N/m2 (or Pa)
Air pressure at sea level is approximately  =  1 atm  =  1 bar  =  105 N/m2  (or Pa)   =  14.50 lb/in2
Water pressure increases 9810 N/m2  for each meter of depth or 0.433 lb/in2 for each foot of depth

Work and Energy or Heat Work  =  Force  x  parallel distance
1 joule, J  =  0.738 ft . lb  =  107 ergs
1 calorie, cal  =  4.18 J  =  0.00442 British Thermal Units, Btu
1 foot pound, ft . lb  =  1.36 J
1 Btu  =  252 cal  =  778 ft·lb  =  1.054 x 103 J
1 electron Volt, eV  =   1.602 x 10-19 J
931.5 million electron volts,MeV is equivalent to 1 atomic mass unit, u

Power  =  Work/time
1 horse power, hp  =  550 ft·lb/s  =  0.746 kilowatts, kW  =  746 W
1 Watt, W  =  1 J/s  =  0.738 ft·lb/s  =  0.00134 hp
1 Btu/hr  =  0.293 W
1 kW  =  1.34 hp

Radioactivity
1 curie  =  3.70 x 1010 disintegrations/second  =  3.70 x 1010 bequerels

<span>SI BASE UNITS<span>Physical Quantity Name of  UnitSymbol</span><span>Lengthmeterm</span><span>Mass kilogramkg</span><span>Timeseconds</span><span>TemperaturekelvinK</span><span>Electric currentampereA</span><span>Luminous intensitycandelacd</span><span>Amount of substancemolemol</span></span>

 

<span>         SI PREFIXES<span>MACROMICRO</span><span>FactorPrefixSymbolFactorPrefixSymbol</span><span><span>1018</span>exaE<span>10-1</span>decid</span><span><span>1015</span>petaP<span>10-2</span>centic</span><span><span>1012</span>teraT<span>10-3</span>millim</span><span><span>109</span>gigaG<span>10-6</span>microμ</span><span><span>106</span>megaM<span>10-9</span>nanon</span><span><span>103</span>kilok<span>10-12</span>picop</span><span><span>102</span>hectoh<span>10-15</span>femtof</span><span><span>101</span>dekada<span>10-18</span>atto<span>a

</span></span></span>
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3 years ago
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