44. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3 , and 1 in.3 = 1.639 10-2 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460ft2 /gal becomes 460 ft2 gal 1m 3.28 ft
2
1 gal 3.79 L
= 11.3 m2 /L .
(
Islamic Final Exam
Revision
Topics to Revise:
A. Sources of Islamic Law
a. Two Cases of Analogy
b. Presumption of Continuity: assuming or continuing the assumption
otherwise (we dont prove original state)
B. Islam and Art:
a. What are the arguments for an
Homework 2
University Physics II
Spring 2016
Name_
-MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A point charge Q of mass 8.50 g hangs from the horizontal ceiling by a light 25.0-cm thread. When
Homework 11
University Physics I
(PHYS 211)
Fall 2015
Name_
-MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A light triangular plate OAB is in a horizontal plane. Three forces, F1 = 6.0 N, F2 = 9.
Homework 12
University Physics I
(PHYS 211)
Fall 2015
Name_
-MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall an
Homework 1
University Physics II-PHYS 212
Spring 2016
Name_
-MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A piece of plastic has a net charge of +2.00 C. How many more protons than electrons doe
Homework 10
University Physics I
(PHYS 211)
Fall 2015
Name_
-MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A turbine blade rotates with angular velocity (t) = 2.00 rad/s- 2.1.00 rad/s3 t2. What i
Homework 9
University Physics I
(PHYS 211)
Fall 2015
Name_
-MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) Two objects are moving in the xy-plane. Object A has a mass of 3.2 kg and has a velocity
5. Various geometric formulas are given in Appendix E. (a) Substituting R = 6.37 106 m (b) The surface area of Earth is 4R2 = 4 6.37 103 km (c) The volume of Earth is 4 3 4 R = 6.37 103 km 3 3
3 2
10-3 km/m = 6.37 103 km
into circumference = 2R
4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain (0.80 cm) (b) and (0.80 cm) 1 inch 2.54 cm 6 picas 1 inch 1.9 picas . 1 inch 2.54 cm 6 picas 1 inch 12 points 1 pica 23 points ,
3. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs = (b) and that distance in chains to be d= (4.0 furlongs)(201.168 m/furlong) = 40 chains . 20.117 m/chain (4.0 furlongs)(201.168 m/furlong) = 160 rods ,
2. The customer expects 20 7056 in3 and receives 20 5826 in3 , the difference being 24600 cubic inches, or 3 1L 2.54 cm 24600 in3 = 403 L 1 inch 1000 cm3 where Appendix D has been used (see also Sample Problem 1-2).
Chapter 10: Parametric Curves and Polar Coordinates
Slope and Tangent Line
Second Derivative
Length of a Parametric Curve
1
Areas of Surfaces of Revolution
Areas in Polar Coordinates
Lengths in Polar Coordinates
2