PHY 138 - Chaper 3 Solutions

PHY 138 - Chaper 3 Solutions - Problem 3.9 For this problem...

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Unformatted text preview: Problem 3.9 For this problem we simply add up the components based on the unit vectors in the south and east directions. First the magnitude based on the distance to the east, d e , and distance to the south, d s : ( ) ( ) m 07 . 8 m 40 . 5 m 00 . 6 2 2 2 = + = + = s s e d d r . The direction can be found from the following diagram Then the angle is = = = 42 m 6.00 m 40 . 5 tan tan 1 1 e s d d and the direction is 42 degrees south of east (132 compass heading). Problem 3.11 (a) Her resultant displacement in terms of base unit vectors east E e and north N e ( ) ( ) ( ) ( ) ( ) N E E N E e e bl 00 . 4 bl 00 . 3 bl 00 . 6 bl 00 . 4 bl 00 . 3 3 2 1 + = + + = + + = e e e d d d d . (b) The total distance she travels is 13.00 blocks d s d e r Problem 3.12 The spelunkers displacement will be expressed in terms of base unit vectors east E e and north N e . The individual displacements are ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) N N E N E E N e d e e e e d e d e d m 150 m 2.5 6 m 08 1 . 30 sin m 125 . 30 cos m 125 m 250 m . 75 4 3 2 1 = + = + = = = And the total displacement then becomes ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) N E N N E E N e e e e e e e d d d d d 5 . 12 m 358 m 150 m 5 . 62 m 108 m 250 m . 75 4 3 2 1 + = + + + + = + + +...
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This note was uploaded on 07/09/2008 for the course PHY 138 taught by Professor Meitzler during the Summer '08 term at Sam Houston State University.

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PHY 138 - Chaper 3 Solutions - Problem 3.9 For this problem...

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