Problem #1
Change the system of the units below into the SI system!(a). 108 km/jam
(b). 1 dyne = 1 gr.cm/s2
(c). 1 slug/kaki3
(d). 1 g/cm3
Answer:
(a). 108 km/jam = (108 km/jam)(1000 m/1 km)(1 jam/3600 s) = 30 m/s
(b) 1 gr.cm/s2 = (1 gr.cm/s2)(1 kg/1000 gr)(1 m/100 cm) = 10-5 kgm/s2 = 10-5 N
(c) karena, 1 slug = 14,59 kg, 1 kaki = 0,3048 m atau 1 kaki3 = 0,02832 m3
1 slug/kaki3 = (1 slug/kaki3)(14,59 kg/1 slug)(1 kaki3/0,02832 m3) = 10-5 kg.m/s2 = 515,2 kg/m3
(d). 1 g/cm3 = (1 g/cm3)(1 kg/1000 gr)(1 cm3/10-6 m3) = 1000 kg/m3
Problem #2
(a) Convert 3598 grams into pounds.
(b) Convert 231 grams into ounces.
(c) How many ng are there in 5.27 x 10−13 kg?
(d) What is 7.86 x 10−2 kL in dL?
(e) A box measures 3.12 ft in length, 0.0455 yd in width and 7.87 inches in height. What is its volume in cubic centimeters?
(f) A block occupies 0.2587 ft3. What is its volume in mm3 ?
(g) If you are going 55 mph, what is your speed in nm per second?
(h) If the density of an object is 2.87 x 10−4 lbs/cubic inch, what is its density in g/mL?
Answer:
(a) 3598 gram = (3598 g)(1lb/454 g) = 7,93 lb
(b) 231 gram = (231 g)(1 lb/454 g)(16 ons/1 lb) = 8.14 ons
(c) 5.27 x 10-13 kg = 5.27 x 10-13 kg (1012 ng/1 kg) = 0.527 ng
(d) 7.86 x 10-2 kL = 7.86 x 10-2 kL (104 dL/1 kL) = 786 dL
(e) you should the volume of a box is calculated thus, V = L x W x h
First you have to convert all the dimensions to the same unit such inches.
L = 3.12 ft(12 in/1 ft) = 37.4 in
W = 0.0455 yd(3 ft/1 yd)(12 in/1 ft) = 1.64 in
Then,
V = 37.4 in x 1,64 in x 7.87 in = 483 in3
Note the question is asking for cm3. We know the conversion from in to cm. We can easly convert in3 to cm3 thus:
(2.54 cm/1 in)3 = (2.543 cm3/1 in3)
Thus, we can convert 483 in3 into cm3 as follows:
V = (483 in3)(2.543 cm3/1 in3) = 7.91 x 103 cm3
(f) 0.2587 ft3 = 0.2587 ft3(123 in3/ft3)(2.543 cm3/1 in3)(103 mm3/1 cm3) = 7.326 x 106 mm3
(g) 55 mi/h = (55 mi/h)(12 in/1 ft)(2.54 cm/1 in)(107 nm/1 cm)(1 h/60 min)(1 min/60 s) = 2.5 x 1010 nm/s
(h) 2.87 x 10−4 lbs/inch3 = (2.87 x 10−4 lbs/inch3)(454 g/lb)(1 in3/2.543 cm3)(1 cm3/1 mL) = 7.95 x 10-3 g/mL
Problem #3
A 10 ft long cuboid bath tub. Water flows into the bath through a faucet with a speed of 0.1 liters/second. If at first the bath is empty, calculate the time needed to fill the tub to the brim!
Answer:
The length of the bath tub s = 10 ft = (10 ft) (0.3048 m/1ft) = 3,048 m.
The bath volume is V = s3 = (3,048 m)3 = 28,31683 m3 = 28316,83 dm3 = 28316,83 L.
Remember that 1 dm3 = 1 L. the time needed to fill the bath until full is
T = 28316.83 L x 0.1 s/L = 2832 s = 47.2 minutes
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