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3 public class Avast { 4 static public void main(String[] scurvy) { 5 Systemoutprint(scurvy[1] + " "); 6 main(scurvy[2]); 7 } 8 public static void main(String dogs) { 9 assert(dogs == null); 10 Systemoutprintln(dogs); 11 } }

Time (s)

Figure 2-14

A High St B West Time (s)

Table 2-1

And, if the code compiles, the command-line invocation:

Position v Time Time t (s) 00 10 20 30 40 50 60 Position d (m) 00 50 100 150 200 250 300



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where Pa is the power available per solid degree Ps is the total radiated power, in watts r is the radius of the sphere (ie, the distance from O to P ) A practical rule of thumb for this problem is to calculate from the surface area of the sphere If you perform the right calculations, you will find that there are approximately 41,253 square degrees on the surface of a sphere By calculating the surface area of the beam front (also in square degrees), you can find the power within that region Now for the matter of gain in a vertical antenna The vertical is not gainless because it does not, in fact, radiate equally well in all directions In fact, the vertical is quite directional, except in the horizontal (azimuth) plane Figure 7-5 shows the radiation pattern of the typical free-space vertical radiator The pattern looks like a giant doughnut in free space (see solid pattern in Fig 7-5) When sliced like a bagel, the pattern is the familiar circular omnidirectional pattern When examined in the vertical plane, however, the plane looks like a sliced figure-8 The gain comes from the fact that energy is not spread over an entire sphere, but rather it is concentrated





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300 250 200 150 100 50 00 Position (m)

10 20 30 40 50 60 Time (s)

What is the result (Choose all that apply) A 1 2 B 2 3 C Compilation fails due to an error on line 4 D Compilation fails due to an error on line 6 E F Compilation fails due to an error on line 8 Compilation fails due to an error on line 9

Figure 2-15 The data table (a), position-time graph (b), and particle model (c) all represent the same moving object

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How long did the runner spend at any location Each position has been linked to a time, but how long did that time last You could say an instant, but how long is that If an instant lasts for any finite amount of time, then the runner would have stayed at the same position during that time, and he would not have been moving However, as he was moving, an instant is not a finite period of time This means that an instant of time lasts zero seconds The symbol d represents the instantaneous position of the runner the position at a particular instant Equivalent representations As shown in Figure 2-15, you now have several different ways to describe motion: words, pictures (or pictorial representations), motion diagrams, data tables, and position-time graphs All of these representations are equivalent That is, they can all contain the same information about the runner s motion However, depending on what you want to find out about an object s motion, some of the representations will be more useful than others In the pages that follow, you will get some practice constructing these equivalent representations and learning which ones are the easiest to use in solving different kinds of problems Considering the motion of multiple objects A position-time graph for two different runners in a race is shown in Example Problem 2 When and where does one runner pass the other First, you need to restate this question in physics terms: At what time do the two objects have the same position You can evaluate this question by identifying the point on the position-time graph at which the lines representing the two objects intersect

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