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22 A 925-kg car moving north at 201 m/s collides with a 1865-kg car moving west at 134 m/s The two cars are stuck together In what direction and at what speed do they move after the collision 23 A 1383-kg car moving south at 112 m/s is struck by a 1732-kg car moving east at 313 m/s The cars are stuck together How fast and in what direction do they move immediately after the collision 24 A stationary billiard ball, with a mass of 017 kg, is struck by an identical ball moving at 40 m/s After the collision, the second ball moves 600 to the left of its original direction The stationary ball moves 300 to the right of the moving ball s original direction What is the velocity of each ball after the collision 25 A 1345-kg car moving east at 157 m/s is struck by a 1923-kg car moving north They are stuck together and move with an initial velocity of 145 m/s at = 635 Was the north-moving car exceeding the 201 m/s speed limit



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Transmission line responses 89 3 From load and line characteristic impedances: (ZL > Zo) VSWR = ZL/Zo (ZL < Zo) VSWR = Zo/ZL 4 From incident (Pi ) and reflected (Pr ) power: 1+ VSWR = 1 Pr/ Pi Pr / Pi [350] [348] [349]

Like linear momentum, angular momentum can be conserved The law of conservation of angular momentum states that if no net external torque acts on an object, then its angular momentum does not change This is represented by the following equation Law of Conservation of Angular Momentum L1 L2

Figure 9-1 When the ice1 skater s arms are extended, the moment of inertia increases and his angular velocity decreases (a) When his arms are closer to his body the moment of inertia decreases and results in an increased angular velocity (b)





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For example, Earth spins on its axis with no external torques Its angular momentum is constant Thus, Earth s angular momentum is conserved As a result, the length of a day does not change A spinning ice-skater also demonstrates conservation of angular momentum Figure 9-11a shows an ice-skater spinning with his arms extended When he pulls in his arms, as shown in Figure 9-11b, he begins spinning faster Without an external torque, his angular momentum does not change; that is, L I is constant Thus, the ice-skater s increased angular velocity must be accompanied by a decreased moment of inertia By pulling his arms close to his body, the ice-skater brings more mass closer to the axis of rotation, thereby decreasing the radius of rotation and decreasing his moment of inertia You can calculate changes in angular velocity using the law of conservation of angular momentum Li thus, Ii

1 package comwickedlysmart2; 2 class Utils { 3 void do1() { Systemoutprint("do1 "); } 4 protected void do2() { Systemoutprint("do2 "); } 5 public void do3() { Systemoutprint("do3 "); } 6 }

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f you observed the movement of a golf ball being hit from a tee, a frog hopping, or a free throw being shot with a basketball, you would notice that all of these objects move through the air along similar paths, as do baseballs, arrows, and bullets Each path is a curve that moves upward for a distance, and then, after a time, turns and moves downward for some distance You may be familiar with this curve, called a parabola, from math class An object shot through the air is called a projectile A projectile can be a football, a bullet, or a drop of water After a projectile is launched, what forces are exerted on the projectile You can draw a free-body diagram of a launched projectile and identify all the forces that are acting on it No matter what the object is, after a projectile has been given an initial thrust, if you ignore air resistance, it moves through the air only under the force of gravity The force of gravity is what causes the object to curve downward in a parabolic flight path Its path through space is called its trajectory If you know the force of the initial thrust on a projectile, you can calculate its trajectory

Recognize that the vertical and horizontal motions of a projectile are independent Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, and then determine the range using the horizontal motion Explain how the trajectory of a projectile depends upon the frame of reference from which it is observed

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5 From reflection coefficient ( ): VSWR = 1+ 1 [351]

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