Monday, January 11th, 2010
TA: Tomoyuki Nakayama
PHY 2048: Physic 1, Discussion Section 3081
Quiz 1 (Homework Set #1)
Name:
UFID:
Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator.
You need to show all of your work for full credit.
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Antarctica is roughly semicircular with a radius of 2000 km. The average thickness of its ice cover is
3000 m.
a) How many cubic centimeters of ice does Antarctica contain?
We first express the radius and thickness in meters:
2000 km = 2000 km × (1000×100 cm)/(1 km) = 2 × 108
cm
3000 m = 3000 m × (100 cm)/(1 m) = 3× 105 cm
The volume of the half cylinder is given by the area of
semicircle times height. Thus we get
V = (πr2/2) × h = [3.14 × (2 × 108)2/2] × 3× 105 =
1.88 × 1022 cm3
b) How many cubic feet of ice does Antarctica contain? (1 ft = 30.48 cm)
We already have the volume in cm3, so all we have to do is to convert it to ft3. Since the original unit
has cm in numerator, we multiply the conversion factor which has cm in the denominator so that they
cancel. We multiply the factor three times. We have
V = 1.88 × 1022 cm3 × (1 ft/30.48 cm)3 = 6.64 × 1017 ft3