Scribe April 25, 2007

On our morning class we started out with a pre-test and right after that, Mr. K puts us in a group to share our answers and hand in one paper. But there wasnt enough time so we continued the group discussion in the afternoon.

And here are the questions & answers on our pretest this morning.

1.The vector represents a velocity of 50 km/h towards [030°]. What is 4v?
(a) 50 km/h towards [-120°] (b) 50 km/h towards [120°]
(c) 200 km/h towards [030°] (d) 200 km/h towards [120°]

-> answer = c

=4x50 =200

the direction stays

2.The airspeed of an airplane is 300 km/h. A wind blows from due east at
50 km/h. In what direction does the plane need to head in order to travel
due north?
(a) [010°] (b) [296°] (c) [304°] (d) [350°]

-> answer = d

bec. if the wind is travelling due to east and the plane wants to go north the angle of the plane should be around 340-360 degrees....

-and the closest one is the 350 deg.

3. Corrie and Allan are pulling on a rock with equal force, they are at an angle
of 40° to each other.
If the resultant vector has a force of 50 newtons, determine the magnitude of the force with which each is pulling.

- its a tail to tail diagram so we use the parallelogram method....

- the answer would be 140, in my diagram I add the 90 & 50 to get the result of the magnitude of the force.....

4. If the resultant force remains at 50 newtons, at what angle could Corrie and
Allan pull in order for Corrie to pull with double the force of Allan?

- the two should pull at a 90 dgrees angle... bec if Allan is pulling 100N & Corrie pulling 50N...the resultant vector will be 50N by using the pythagorean theorem....

5. An airplane travels at a speed of 250 km/h in a direction 30° north of west for
one hour. It then travels at a speed of 300 km/h for 40 minutes in a direction
of 42° east of north. To return directly to its starting point, in what directionand what distance must the airplane travel?

-first change the velocity of the plane to distance...

250 x (60/60) ->60mins represents 1hr
= 250km

300 x (40/60)

= 200km

- then punch in all the info into the calculator and use the trisolv2 program

-and the result would be like this

-The plane should head of Bearing 163.187 deg. (73.187 + 90) with a distance of 285.848 km.

I gues thats it for me.....

The next scribe will be Al Andrei.............