Physics - Vector Quiz

Solutions to, for rewriting

 

The following are the problems worked out which were not finished in class due to time constraints.  Use them to work thru the problems you might have missed on the quiz.

If you took the late quiz.

You must do the rewrite for that quiz, which will follow.

However I believe a review of both would be in order if you attained any

score lower than 50 on the problem section of either quiz.

 

32.           First record each of the motions in an east west and a north south summation:

 

                                N-S (blocks)                  E-W (blocks)

                                +12                -8                             +7                -5

                                                -3                             +2

                                                -3

                SUM:                +12                -14  =  -2                +9                -5  =                  4             

 

                thus                South 2 blks                         East 4 blks

 

                From Pathagorean theorm 4.47 blocks and -tan 2/4 gives an angle of 26.6o S of E.

 

33.           From Pathagorean theorm 5.8 M/Sec and from the tangent funcion 31o forward of the motion of the belt.

 

34.           Again from Pathagorean theorm 5.5 M and from the tangent funcion 53.1o South of West.

 

35.           From Sin 35o * 7500 N = 4302 N West

                From Cos 3535o * 7500 N = 6143 N North

 

36.           First each leg of the trip is to be resolved into distances, (from the speed formula (last section this is 25 Km each)and then resolved into its components:

 

                Sin (&  Cos) 45o * 25 Km = 17.7 Km both south and west.

                Sin 30o * 25 Km = 12.5 Km north and Cos 30o * 25 Km = 21.7 Km west.

                For a total of 5.2 Km south and 39.4 west.

                From Pathagorean theorm distance is 39.7 Km and the angle is 1.45o South of West.

 

37.                 Resolving the wind yields 21.2 Km/Hr both South and East.  Adding this to the original 80 Km per hour north gives 58.8 Km/Hr north and 21.2 Km/Hr east.  The Theorm of pathagoreas gives 62.5 Km/Hr as the speed and the inverse tangent of 21.2/58.8 tells us the angle is 19.8o North of East

 

38.           Note I will be using [ x ] tko imply the square root of x as i cannot make the radical on this system.

 

                Td = [ 2 * 25 M / 9.8 M/Sec2 ] = 2.26 sec                                          thus

                1100M / 2.26 sec  =  486.99 M/sec

 

39.           Sin 60o * 300 M/sec gives an initial vertical speed of 259.8 M/sec.

                259.8 M/sec  /  9.8 M/sec2  = 26.5 sec (UP) * 2 = 53.0 seconds aloft.

 

40.           We are given both a horizontal and vertical initial speed of 200 M/sec thus

                200M/sec  /  9.8M/sec2  =  20.45 seconds going up, which also implies a total time aloft of 40.9 seconds.  This time is spent traveling at 200 M/sec over the water for a distance of 8180 M.

 

NOTICE

 

          Many of these kproblems wer worked out without showing the formula substatutions.  For the rewrite is is necessary for you to include the setup and canceling on your paper.