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which, since V 0 , I 0 , and are all constants in any given case, shows that p is also constant in any given case in any balanced three-phase system. Problem 206 Show that p, in eq. (445), is EQUAL to PT in eqs. (440) and (444).

2. Given the following,

11. 12. 13. 14. 15. String x = "xyz"; x.toUpperCase(); String y = x.replace('Y', 'y'); y = y + "abc"; System.out.println(y);

Let us begin by observing that a SINGLE plane vector is de ned in terms of two independent variables, its MAGNITUDE and its ANGULAR POSITION relative to an agreed-upon reference axis. The independent variables are also referred to as degrees of freedom ; thus, a single " plane vector is said to have two degrees of freedom. Such a vector, A A=h, is illustrated in Fig. 259, where A and h are the two degrees of freedom. (As always, positive angles are measured in the ccw direction from the reference axis.)

3. Given the following,

-4.00 0 -720

13. 14. 15. String x = new String("xyz"); y = "abc"; x = x + y;

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Next consider a balanced set of three plane vectors. As we know, this is any set of three vectors having EQUAL MAGNITUDES and EQUAL PHASE DISPLACEMENTS. " " " Now let A1 , B1 , and C1 be such a balanced set, in which the equal phase displacement is "1 as the reference vector, displaced an angle h from the reference 1208, and let us take A " " " axis, as illustrated in Fig. 260, where jA1 j jB1 j jC1 j A1 B1 C1 : We are already familiar with the fact that the vector sum of such a balanced set of vectors is equal to zero. In this regard, note that a balanced set of plane vectors has just TWO degrees of freedom, these being the common magnitudes of the vectors and the angular displacement h of the reference vector from the reference axis. Thus the common magnitude, A1 B1 C1 , and the reference angle h are the two degrees of freedom in Fig. 260.

A. 2 B. 3 C. 4 D. 5 4. Given the following,

Next, in Fig. 260, assuming the diagram to be drawn on the complex plane, note that* " " B1 A1 j120 and " " C1 A1 j240

14. 15. 16. 17. 18. String a = "newspaper"; a = a.substring(5,7); char b = a.charAt(1); a = a + b; System.out.println(a);

5. Given the following,

where 120 and 240 are understood to be angles in degrees{ (1208 and 2408), and thus that general form of the algebraic equation for the balanced case of Fig. 260 can be written as " " " A1 A1 j120 A1 j240 0 446

2.00m

4. 5. 6. 7. 8. String d = "bookkeeper"; d.substring(1,7); d = "w" + d; d.append("woo"); System.out.println(d);

Using the java.lang.Math Class (Exam Objective 8.1)

the right-hand side re ecting the fact that the vector sum of such a balanced set of vectors is zero. " " " Now consider an unbalanced set of three plane vectors A, B, and C, such as is illustrated in Fig. 261, in which let " " " " S A B C 447

6. Given the following,

" " * Let V V ja be a vector quantity on the complex plane. Now multiply V by jb , thus " V jb V ja jb V j a b " " " showing that multiplying a vector V by rotates V through the angle b but does not change the magnitude of V .

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. public class Example { public static void main(String [] args) { double values[] = {-2.3, -1.0, 0.25, 4}; int cnt = 0; for (int x=0; x < values.length; x++) { if (Math.round(values[x] + .5) == Math.ceil(values[x])) { ++cnt; } } System.out.println("same results " + cnt + " time(s)"); } }

6.00m

A. same results 0 time(s) B. same results 2 time(s) C. same results 4 time(s) D. Compilation fails. E. An exception is thrown at runtime. 7. Which of the following are valid calls to Math.max (Choose all that apply.) (Yeah, yeah, we know that on the real exam you d know how many were correct, but we just want you to work a little harder here.) A. Math.max(1,4) B. Math.max(2.3, 5)

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