11 Pages

C_NOTE2

Course: PRG 155, Fall 2009
School: CSU Fullerton
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Programming C -Topic 2 Revision 2 Sep 2000 Programming in C Topic 2 - Data Types and Declarations 1.0 VALUES AND VARIABLES 1.1 The decimal number system is based on ten number symbols, 0 through 9. In decimal, the first (rightmost) position tells how many ones, the second (to the left) tells how many tens, and so forth. The value twenty-four is two tens and four ones, written 24. Similarly, 4680 means four...

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Programming C -Topic 2 Revision 2 Sep 2000 Programming in C Topic 2 - Data Types and Declarations 1.0 VALUES AND VARIABLES 1.1 The decimal number system is based on ten number symbols, 0 through 9. In decimal, the first (rightmost) position tells how many ones, the second (to the left) tells how many tens, and so forth. The value twenty-four is two tens and four ones, written 24. Similarly, 4680 means four thousands, six hundreds, eight tens, and no ones. 2.0 BINARY NUMBERS 2.1 The computer's transistors acting as switches have only two possible states, off and on. Therefore, the computer is capable of working only with those states. We symbolize them with the digits zero and one. This means that instead of using the decimal system with ten digits, the computer must express numbers in a binary system-one with only two symbols, 0 and 1. Though humans express the value nine as 9, the computer must express it with a series of zeros and ones, 1001. 2.2 The decimal number 712048 7 ---- hundred thousands 1 ---- ten thousands 2 ---- thousands 0 ---- hundreds 4 ---- tens 8 ---- ones 2.3 The binary number 1010111 (decimal 87) 1 ---- sixty-fours 2^6 0 ---- thirty-twos 2^5 1 ---- sixteens 2^4 0 ---- eights 2^3 1 ---- fours 2^2 1 ---- twos 2^1 1 ---- ones 2^0 2.3.1 Binary 1001101 1 * 64 = 64 + 0 * 32 = 0 + 0 * 16 = 0 + 1*8=8 + 1*4=4 + 0*2=0 + 1*1=1 = 77 Decimal 77 3.0 HEXADECIMAL NUMBERS 3.1 Hexadecimal numbers, are numbers with the base sixteen, because one hexadecimal digit represent exactly four binary digits. 1 C Programming -Topic 2 Revision 2 Sep 2000 3.2 A base-sixteen number requires sixteen number symbols, so we borrow 0 through 9 from the decimal system and add A, B, C, D, E, and F for the values above 9 (see table). Decimal 10 in hexadecimal ("hex" for short), then, is A, and F is decimal 15. Decimal 16 would be hex 10. 3.3 To convert from binary to hex, group the binary digits in fours beginning from the right and convert each group to the appropriate hex digit. For example, we would separate binary 1101100 as 110 1100. The first group is decimal 6 or hex 6, and the second is decimal 12 or hex C. The complete hex number is 6C. 3.4 HEXADECIMAL 4BD2 is decimal 19410 e.g 4 B D 2 4 11 13 2 * * * * 4096 256 16 1 = = = = 16384+2816+208+2 = 19410 3.5 Decimal 19410 16^3 + 16^2 + 16^1 + 16^0 3.6 Number System Conversion Decimal Binary Octal 0 0 0 1 1 1 2 10 2 3 11 3 4 100 4 5 101 5 6 110 6 7 111 7 8 1000 10 9 1001 11 10 1010 12 11 1011 13 12 1100 14 13 1101 15 14 1110 16 15 1111 17 16 10000 20 17 10001 21 18 10010 22 19 10011 23 20 10100 24 Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 * In ANSI C, we precede the hex number with 0x. For example, decimal 46 would be hex 0x2E; decimal 25 would be hex 0x19. 4.0 OCTAL NUMBERS 4.1 Conversion between octal and decimal follows the same processes as between hex and decimal except that you are dealing with eight number symbols and powers of eight-1, 8, 64, 512, and so forth. In ANSI representation, the octal number is preceded by a zero. 5.0 CHARACTER VALUES 5.1 When we looked at number systems we saw three different ways of expressing numbers in C-decimal, octal, and hex. There is a fourth, ASCII ( or EBCDIC if we happened to be using IBM mainframe type of computer). By enclosing a character in single quotes (apostrophes) we are really referring to its ASCII code. The notation 'A' is the same as 65 or 0101 or 0x41. The 2 C Programming -Topic 2 Revision 2 Sep 2000 notation ' '(a space) is the same as 32 or 040 or 0x20. We could add, for instance, 32 + 65 or space) is the same as 32 or 040 or 0x20. We could add, for instance, 32+65 or ' '+0x41 or ' '+'A' or any combination of representations and, depending on how we wish to display the result, show 97 or 0141 or 0x61 or the character a (whose ASCII code is 1100001 or decimal 97). Remember, once it's stored, it's just a set of bits. 5.2 Some Characters in either the ASCII or EBCDIC codes are not on the keyboard and so would be difficult to show in character notation. An example might be the form-feed code sent to a printer to tell it to go to the next page. In C we represent these characters with special characters, each of which consists of a backslash followed by a printable character. Even though two characters are shown within the quotes, it represents, and is stored as, just a single character. Following are the list: \a \n \t \ Audible alarm Newline Horizontal tab Double quote \f \r \ \\ Form feed Carriage return Apostrophe (single quote) Backslash 6.0 Variables and Variable Names: 6.1 A variable represents a space in the computers main memory. We put values in that memory location for use later in the program. 6.2 We can use many characters in a variable name. How many depends on the compiler. But ANSI recommends at least 31 will be significant. PreANSI specifies only 8. 6.3 We can use only alpha characters (A to Z or a to z), numeric characters (0 to9), or under score (_) in variable name. A space in variable name is not allowed. 6.4 Variable name should start with an alpha character or an underscore, not a number. 6.5 C program is case sensitive. GAME, Game, and game are all different. BY tradition the variable names are all lower case 6.6 There are 32 reserved words (a special meaning to compiler) which should not be used as variable names. They can be used only in program statements. Some examples are: char, do, else, float, for, if, int, long, short, signed, sizeof, void, while 3 C Programming -Topic 2 Revision 2 Sep 2000 Topic 2 (continued) Data Types and Declarations 1.0 Declarations and initialization/ assignment: 1.1 A declaration performs a number of functions: 1. It directs C as to how to store the value, whether it can have a fractional component or whether it is allowed to be negative as well as positive. 2. It automatically defines it, meaning that it allocates memory to the value or variable. The type of declaration tells the computer how much memory. 3. It may also initialize a variable-assign a meaningful value to that space in memory. 1.2 Declarations follow this general form (the stuff in brackets, [], is optional); data type variable[ = initialization]; 1.3 For example: int frequency; int StartValue = 14; 1.3.1 Notice the semicolons at the end of the declarations; they are statements and must end in semicolons. 1.3.2 We can declare more than one variable of a single type in one declaration statement by separating the declarations with commas. int total, counter, interval; int BeginRange = 0, EndRange = 100; 1.3.3 We can mix initialized and non initialized declarations in the same statement (as long as they are of one type), but many consider it improper form. Notice that in the following declaration, EndRange is initialized to 100, but BeginRange is not initialized at all. int BeginRange, EndRange = 100; 2.0 INTEGRAL DATA TYPES 2.1 Integral data types allow only integer, or whole, numbers-no decimal points. There are two basic integral types: char for "character" and int for "integer." The char data type is eight bits long. We can also store numbers there. 2.2 In a single char space we can store the code for A or perhaps the number 65. Since the ASCII code for A consists of the same bits as the binary representation for decimal 65, exactly the same value, 01000001, will be stored. Note the leading zero bit to make up eight bits. 2.3 Either integral data type may have one of two modifiers, signed or unsigned. The signed modifier means that the value may be either positive or negative. 2.3.1 Under ANSI rules this is the default, so if there is no modifier, the char (or int) is assumed to be signed. The unsigned modifier means that the number may only be positive. An unsigned char may have values from 0 to 255, while (because it takes up one bit for the sign) signed chars may have values ranging from -128 to 127. By convention, we do not use signs with octal, hex, or character notation (such as 'A'); they may only be positive. 2.4 Some valid char declarations are: char letter; unsigned char index; char TopGrade = 'A', BottomGrade = 'F'; char TopGrade = 65, BottomGrade = 70; 2.5 Most C compilers for PCs use a 16-bit int. (2 bytes) 2.6 The int data types may have a signed or unsigned modifier but it may also have short or long modifiers. ANSI tells us only that a long int will have at least as many bits as a short int but in most implementations a short int is 16 bits while a long int is 32. The default, if neither short 4 C Programming -Topic 2 Revision 2 Sep 2000 nor long is specified, is either a short or long integer, depending on the word size of the machine. 2.7 The data type int is the default. If a declaration has modifiers but no data type, it is assumed that it is an int of some kind. 2.8 Some valid int declarations are: int current_page, last_page; short age; /* Equivalent to short int age*/ unsigned volts=110; /*Equivalent to unsigned int volt=110*/ unsigned volts='n'; /*Equivalent to declaration above */ unsigned long NationalDebt; signed short Variance; /*Modifier signed not needed under ANSI */ 2.9 Unless otherwise stated, integral values default to type int (remember, that may be the same size as either short or long depending on your machine). Decimal-notation, integers and character values are stored as signed (even though we would never state a character such as A' with a negative sign), and octal and hex as unsigned. 2.10 In any case, if the value exceeds the size of a short int, it will be stored as a long int. For example, if the default int on your machine is 16 bits (with a maximum decimal value of 32767) and you put the number 145832 in your program code, the compiler will store it in a long int of 32bits. 2.11 Notice that character values, even though they fit within 8 bits, the size of data type char, are stored as ints. 2.12 We may also explicitly declare values by following the value with an L or a U (or l or u) or both. The postfix L forces the value to be stored as a long and u as unsigned. The value 56ul would be stored as an unsigned long. 3.0 FLOATING-POINT DATA TYPES 3.1 In scientific notation, we use a mantissa of significant digits multiplied by some power of ten. For example, 7146 might be expressed as 7.146 X 10^3. The two values are the same; 10^3 is 1000, and 7.146 X 1000 is 7146. In our C program we can write the number as 7146.0 or 7.146E3, that is, a mantissa of 7.146 and a power-of-ten exponent of 3. The second form is referred to as E notation. 4.0 ANSI C DATA TYPES 4.1 E.g. Integral Signed: char short Int [4 or A'] long [4L] Unsigned: unsigned char unsigned short unsigned [4U] unsigned [4UL] Floating Point Signed: float [4.2F] double [4.2] Long double [4.2L] 4.2 There are three different floating-point data types: float, double, and long double. In terms of the size of each, all ANSI guarantees is that a double is greater than or equal to a float and a long double is greater than or equal to a double. In many microcomputer Cs, a float is 32 bits, a 5 C Programming -Topic 2 Revision 2 Sep 2000 double is 64, and a long double 80. Unlike integral data types, floating-point data types always allow positive or negative values, and so cannot have the modifiers signed or unsigned. 4.3 Following are some examples of floating-point-variable declarations: float WageRate = 12.75; double Area, Volume; long double humongous; 4.4 When declaring a long double, be sure to state long double, not just long, because that would default to a long int. 4.5 A value with a decimal point will normally be stored as a double. We may explicitly declare it as a float with F or f, or a long double with L l. A number written as 3.806E-3F (or .003806F) would be stored as a float instead of a double. A value with a large number of significant digits or a large exponent will be stored as a long double. 5.0 STRING DATA 5.1 A string is a set of characters. We include string values in a program by enclosing the characters in quotes; for example: "This is a bunch of characters" 5.1.1 The quotes are not part of the string; they just serve to tell the C compiler where the string begins and ends. To include quotes as part of the string we must use the special character \". This is true of any of the special characters except the apostrophe and question mark, although the backslashes also work there. 5.1.2 "He said \ "I\'m drowning in a sea of C!\" " Would be stored as He said "I'm drowning in a sea of C!" /* \'not necessary */ 5.2 The compiler will concatenate (connect together end to end) adjacent strings. For example "This is just" "one string." will be stored as This is just one string. 5.2.1 Remember, white space, such as spaces, tabs, and line endings, is ignored by the compiler (unless it is inside quotes), so this concatenation is property often used to write a single long string using two lines in C code. "This is just" "one string." is stored the same. 5.2.2 Since we can't put line endings between quotes, the following wouldn't work. "This is just one string." 5.3 There are no string variables in C. 6..0 ARITHMETIC EXPRESSIONS 6.1 An expression is anything that can be reduced to a single value. 6.2 Expressions that require some evaluation by the computer-they are made up of more than one value or variable, for example, 26 + 17. 6.3 An arithmetic expression consists of values and/ or variable connected by arithmetic operators, which tell the computer how to combine the values. The expression 26 + 17, for example, uses an operator, +, indicating that the values on either side, 26 and 17, are to be added together. 6.4 There are strict rules about which operation is to be done first. The rules involve 6 C Programming -Topic 2 Revision 2 Sep 2000 precedence, a hierarchy of operations that dictate the types of operations that are to be performed before other types; and associativity, which dictates order if two operations have the same level of precedence. 6.4.1 Table below shows the arithmetic operators in precedence, highest first, and their associativity. 6.5 We can force calculations to be in any order we choose by enclosing some of them in parentheses. 6.6 Arithmetic Operators in Precedence Level 1 2 3 4 4 5 Type Unary Cast Multiplicative Additive Assignment Associatively Right to left Right to left Left to right Left to right Right to left Operator Negate Plus Size A data type Multiply Divide Remainder Add Subtract Equals Accumulation Symbol + sizeof (type) * / % + = +=, -=, *=, /=, %= Example -4 +4 sizeof x, sizeof (int) (int),(float),etc. 6*4 6/4 6%4 6+4 6-4 x=4 x += 16 x %= 4 6.7 Handwritten Expression 6+4 -------- * 4 2 -------------- = 2 3*2 4 + ----------2-1 6.7.1 In C Without Parentheses 6+4/2*4/4+3*2/2-1 \ \ / / / \ / / / \ 2 / / 6 / / \ \ / / \ / / \ 8 / 3 / \ \ / / / \ 2 / / \ / / / 8 / / \ / / \ / / 11 / \ / = 10 In C With Parentheses ((6+4) / 2 * 4) / (4 + (3 * 2 / (2 -1)) \ / / / \ / \ / 10 / / \ 6 1 \ / / \ \ / 5 / \ 6 \ / \ / 20 10 \ / \ / \ / \ / \ / \ / =2 6.7.2 The parentheses are, indeed, unnecessary, but if they improve the readability or understandability of the program, they can certainly be included. 7 C Programming -Topic 2 Revision 2 Sep 2000 7.0 INTEGER ARITHMETIC 7.1 If the data types in a particular operation are integer then the result will also be integer. This typically causes no problem except when the result exceeds the maximum capacity of an integer or when you are dividing. If you exceed the limits of an integer you will get a meaningless result. For example, 32000 * 10 (in a C that defaults to a short int) yields the result -7680. 7.2 When dividing an integer by an integer the result will be an integer. The result of the expression 3 / 2 is 1, not 1.5. Since both 3 and 2 are integers (they have no decimal points) an integer calculation will be done yielding an integer result. 7.2.1 The result of the expression 5 % 3 is 2 because 5 divided by 3 is one with a remainder of 2. As the following long division shows, 762 % 35 is 27. 7.2.2 Be careful when using the remainder operator with negative numbers; the results vary with the particular implementation of C. 8.0 MIXED ARITHMETIC 8.1 In general, when data types are mixed in an expression, each operation will be performed at the highest data type involved in the expression-highest meaning the one that takes up the largest amount of memory. Since there is some overlap, floating-point types are considered higher than integral. 8.2 It is important to recognize that the calculations are not all performed at the highest data type included in the expression; each operation is evaluated separately and performed at the highest data type involved in just that operation. Eventually the result will be of the highest data type in the entire expression but it may take a while to get there. 8.3 For example: 8.3 + 5 / 2 [=10.3] \ \ / \ 2 (int) \ / 10.3 (float) 8.3.1 The expression 5 / 2 was performed first and since both the 5 and the 2 are integers (they have no decimal points) 5 / 2 was performed as type int with a result of 2 (not 2.5). 8.4 Compare that with this: 8.3 + 5 / 2 [=10.8] \ \ / \ 2.5 (double) \ / 10.8 (double) 8.5 What does all this mean to you? Computing is a series of trade-offs. If you declare variables as type float rather than double (perhaps to save memory space), calculations on these floats will be done as double. In other words, the computer will have to go through extra conversions-floats to doubles and then back to float for the result. The trade-off is storage space versus execution speed. 9.0 CASTS 9.1 You don't have to be satisfied with the data type that C determines for an expression; you can change the data type to anything you want by casting it. The cast operators are simply the desired data-type key word enclosed in parentheses. For example: (int) (long double) (unsigned short) (float) 9.2 They may not look like it but they are operators, like + or - or *. They are unary operators 8 C Programming -Topic 2 Revision 2 Sep 2000 with right-to-left associatively, but fall on the precedence level below negate and plus9.3 Given these initialized declarations: int x = 5, y = 2, z = 3; the following expressions will evaluate as shown: x / y * z [=6] x / y * (float)z [=6.0] x / (float)y * z [=7.5] 9.3.1 The resultant value of a floating-point number cast as an integer follows the same rules as an integer division-only the whole-number part of the value survives. If your program contained the declaration float x = 2.7;, the value of the expression (int)x would be 2. Note that the value of the variable x would not be changed; it would still be 2.7. Like any other operator, the cast operator only uses the values around it; it does not change them any more than 4 + x would change the value of x. 9.4 You must be careful with your casts. If you cast a value as a smaller data type, a long as a short, for example, C will just copy as many bits as it can fit and the result will be unrecognizable. The value of (short)45612 in a typical C implementation is -19924. 10.0 THE sizeof OPERATOR 10.1 The result of the sizeof operator is an integer equivalent to the number of bytes required to store the object following it. This object can be either a data type or an expression. sizeof (data type) or sizeof expression 10.2 The sizeof operator is unary and has the same precedence and associatively as the other unary operators. If the object of sizeof is a data type, then the data type must be in parentheses. 10.3 As we saw earlier, various Cs have different storage requirements for the same data types. The expression sizeof (int) typically yields 2 or 4 depending on the C. 10.4 One way to find out the sizes of the various types would be to print out the sizeof each. sizeof (char) should produce 1 in every C. 10.5 The sizeof operator can be used in expressions just like any other unary operator. For example, given these declarations and initializations: int a = 10; /* Assume a 2 - byte int */ float b = 20.5; /* Assume a 4 - byte float */ this expression, sizeof a + b would evaluate to the float value 22.5 (the sizeof a, 2 plus 20.5). Remember, sizeof is unary and higher in precedence than addition. sizeof (a + b) would evaluate to the integer value 4. Since the result of 10 + 22.5 would be float, sizeof would yield the number of bytes in the data type float. 11.0 ASSIGNMENT 11.1 Variables identify spaces in main memory. These spaces are variable because they can contain various and changeable v...

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Introduction to Natural Language Processing with NLTKSteven Bird Edward Loper Ewan KleinUniversity of Melbourne, AUSTRALIA University of Pennsylvania, USA University of Edinburgh, UKminor edits by Anoop Sarkar any mistakes are his faultPython
Sveriges lantbruksuniversitet - CMPT - 825
Homework #3: CMPT-825Anoop Sarkar anoop@cs.sfu.ca(1) Part-of-speech Tagging: Consider the task of assigning the most likely part of speech tag to each word in an input sentence. We want to get the best (or most likely) tag sequence as dened by th
Allan Hancock College - MOPGPR - 20022002
MEMBERS OF PARLIAMENT (LIFE GOLD PASS) REGULATIONS 2002 2002 NO. 313 MEMBERS OF PARLIAMENT (LIFE GOLD PASS) REGULATIONS 2002 2002 NO. 313 - TABLE OF PROVISIONSPART 1-PRELIMINARY 1. Name of Regulations 2. Commencement 3. Definitions
Sveriges lantbruksuniversitet - CMPT - 825
CMPT 825 Natural Language ProcessingAnoop Sarkarhttp:/www.cs.sfu.ca/~anoopGoals of the Course Convince you that understanding language is a subtle, interesting, and tractable problem Give insight into various algorithms including statistical ma
Neumont - EN - 1973
Supreme Court of Canada Association of Radio and Television Employees of Canada (CUPE-CLC) v. Canadian Broadcasting Corporation, [1975] S.C.R. 118 Date: 1973-10-02 Association of Radio and Television Employees of Canada (CUPE-CLC) Appellant; and Cana
Sveriges lantbruksuniversitet - CMPT - 413
Homework #8: CMPT-413Distributed on Mon, Apr 4, Due on Mon, Apr 11Anoop Sarkar anoop@cs.sfu.ca(1)Implement the feature unification algorithm given in Figure 11.8 in Jurafsky and Martin (on page 423). Download the file FeatureStruct-inc.pm whic
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CMPT 880 Fall 2007 Presentation Summary & Question about BTName: Michael Jia Email: zmj@sfu.ca Student number: 301046057Towards a Global IP Anycast ServiceIP anycast was first proposed in 1993 and now became an important IP packet addressing and
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Chapter7SynchronizationTopics Physicalclocksynchronization Logicalclocksynchronization Causalityrelation Lamportslogicalclock Vectorlogicalclock Multicast ISISvectorclockSnapshotNewIssuesinDS Globaltime Eventordere1at1
CSU Fullerton - BTO - 120
# classes script - place a list of students into course sectionssection=Awhile read inputdo echo "BTO120$section $input" > /tmp/classes.temp.$ section=$(echo $section | tr [A-N] [B-NA])done < classlist.txtsort -k1,1 -k3 /tmp/classes.temp.
Sveriges lantbruksuniversitet - STAT - 201
Review for Final Exam Course Coverage: Chapters 1 through 20. Excluded sections and topics: Ch 1: pie charts, bar graphs, stemplots, time plots. Ch 2: boxplots, computing formulas for quartiles. Ch 6: conditional distributions. Ch 7: using random num
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General Knowledge Questions from Old Exams Set III. General knowledge questions. (5 points each.) From April 2007. III.1. A circle with diameter AP intersects a circle with diameter BP at P and a second point Q. Prove that Q always lies on the line A
Laurentian - MATH - 103
MATH 103 200710 Problem Set 4Edward Doolittle Thursday, February 22, 2007The following problems may appear on the quiz on Thursday, March 1, 2007. 1. Find the stationary points of the following functions:2. For each of the functions in the previo
Laurentian - CS - 330
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Sveriges lantbruksuniversitet - ARTS - 19981
PROFILE OF STUDENTS IN SFU COURSES COURSE: CNS 380-3 H01 LOCATION: DOW TITLE: STT-CAN.POL.ECONOMY SECTION TYPE: SEC SEMESTER: 1998-1 ENROL: 16
Sveriges lantbruksuniversitet - POL - 19981
PROFILE OF STUDENTS IN SFU COURSES COURSE: POL 447-3 D02 LOCATION: SFU TITLE: INTER. POL. ECONOMY SECTION TYPE: SEM SEMESTER: 1998-1 ENROL: 14
Sveriges lantbruksuniversitet - APSC - 19983
Sheet1 PROFILE OF STUDENTS IN SFU COURSES COURSE: CMNS 240-3 D01 LOCATION: DOW TITLE: POLITICAL ECONOMY SECTION TYPE: LEC SEMESTER: 1998-3 ENROL: 36 = PROGRAM OF STUDENT (Top 5 programs reported in each category Programs with < 3 students not shown s
Sveriges lantbruksuniversitet - APSC - 20031
Sheet1 PROFILE OF STUDENTS IN SFU COURSES COURSE: CMNS 240-3 ALL SECTIONS LOCATION: SFU TITLE: POLITICAL ECONOMY SECTION TYPE: LEC SEMESTER: 2003-1 ENROL: 71 = PROGRAM OF STUDENT (Top 5 programs reported in each category Programs with < 3 students no
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Sheet1 PROFILE OF STUDENTS IN SFU COURSES COURSE: POL 447-4 D01 LOCATION: SFU TITLE: INTL.POL.ECONOMY SECTION TYPE: SEM SEMESTER: 2000-3 ENROL: 17 = PROGRAM OF STUDENT (Top 5 programs reported in each category Programs with < 3 students not shown sep
Sveriges lantbruksuniversitet - POL - 20033
Sheet1 PROFILE OF STUDENTS IN SFU COURSES COURSE: POL 447-4 D01 LOCATION: SFU TITLE: INTL.POL.ECONOMY SECTION TYPE: SEM SEMESTER: 2003-3 ENROL: 16 = PROGRAM OF STUDENT (Top 5 programs reported in each category Programs with < 3 students not shown sep
Allan Hancock College - SAA - 200536
Western Australia Stamp Amendment (Assessment) Act 2005 Western Australia Stamp Amendment (Assessment) Act 2005 CONTENTS 1. Short tit
Sveriges lantbruksuniversitet - POL - 20033
Sheet1 PROFILE OF STUDENTS IN SFU COURSES COURSE: POL 356-4 D01 LOCATION: SFU TITLE: POL.ECONOMY/LABOUR SECTION TYPE: SEM SEMESTER: 2003-3 ENROL: 27 = PROGRAM OF STUDENT (Top 5 programs reported in each category Programs with < 3 students not shown s
Toledo - PHY - 110
35.SSM REASONING AND SOLUTION According to Equation 3.3b, the acceleration of the astronaut is a y = (vy - v0y )/ t = vy / t . The apparent weight and the true weight of the astronaut are related according to Equation 4.6. Direct substitution gives
Toledo - PHY - 110
6.REASONINIG AND SOLUTION Using Equation 28.1 with v 1v 1 1 2 c c we have2 1v t0 = t 1 2 c 22The difference between t and t0 is v Time difference = t t0 = t ( 1 ) 2 c2 8.64 104 s 1 7800 m/s 2 4 Time difference = (15 day
Toledo - PHY - 110
REASONING AND SOLUTION From Equation 16.1, we have O = v/f. But v = x/t, so we find v x 2.5 m = = 0.49 m O= = f t f (1.7 s )( 3.0 Hz ) _ 5.13.SSMREASONINGThe tension F in the violin string can be found by solvingEquation 16.2 for F to obtai