# READ ALL CHECKPOINTS IN THE CHAPTER !
# STUDY THE NUMBERED OR UNNUMBERED FIGURES, GRAPHS, AND TABLES.
# IT DOES NOT HURT TO DO EXTRA END-OF-CHAPTER REVIEW QUESTIONS, PLUG AND CHUGS, RANKINGS , OR EXERCISES AND PROBLEMS.
#IF TIME PERMITS GET MORE PRACTICE WITH THE OUT OF SEQUENCE SAMPLE MULTIPLE CHOICE QUIZZES AT VERTICAL LEFT HAND COLUMN Q1, Q2, Q3, LINKS ETC HERE~ http://www.nvaphysics.com/P11AU11.htm .
|Quiz 5 (These problems will contribute to real quiz 3 next Tuesday) Physics 11 Autumn 2011 --- Selected topics + exercises/problems from Ch. 5|
|Quiz 5 Physics 11 Autumn 2011 --- Selected topics from Ch. 5|
|REVIEW QUESTIONS---1, 4, 5, 10, 11, 12, 13, 14, 15, 17, 18, 19, 21, 22|
|PLUG AND CHUG---1, 2, 3, 4,|
|RANKING--1, 2, 3|
|EXERCISES--1, 2, 3, 5, 6, 7, 14, 15, 16, 20, 21, 22, 23, 24, 27, 31, 32, 33, 38,|
|PROBLEMS--1, 2, 3, 4, 5, 6|
|DISCUSSIONS ---CHECK BACK LATER !|
|1. Your push on the wall is the action; the wall exerts a reaction on you. See the push boarding in class quiz. .|
|4. page 68.|
|5. See figure 5.4|
|10. Figure 5.14|
|11. The larger the mass, the smaller the acceleration for the same force magnitude. Which has the largest mass, cannonball or cannon? SEE FIGURE 5.16 AND RELATED DISCUSSIONS|
|12. Figure 5.7, page 68.|
|13. If the helicopter exerts a downward force on the air, the air exerts a force of the same magnitude and opposite direction on the copter.|
|14. Between two objects, for every action force there is a reaction force.|
|15. See page 73 summary of the three laws.|
|17. Vector example is velocity; name two more. Scalar example is temperature; name two more.|
|18. Does speed have a direction?|
|19. See figure 5.27.|
|21. The net vertical component of the force of string on Nellie must balance her weight. See the parallelogram in figure 5.25.|
|22. Figure 5.27.|
|PLUG AND CHUG|
|1. JUST SUBTRACT THE MAGNITUDES TO GET THE RESULTANT MAGNITUDE. The direction of the resultant must be North.|
|2. Figure 5.23, figure 5.24|
|3. Figure 5.23, figure 5.24|
|4. Figure 5.23, figure 5.24. USE PYTHAGOREAN THEOREM.|
|1. Use the parallelogram rule of figure 5.27. B has the largest resultant magnitude|
|2. a has the greatest vertical component; a and b have the SAME horizontal component. SEE FIGURE 5.25 in which THE RIGHT ROPE has THE GREATEST TENSION; generally the rope that has the largest angle with the horizontal has the largest tension.|
|3. For each figure, draw a parallelogram from the two vectors; the diagonal is the resultant. In A, the parallelogram is a simple rectangle since the vectors are at right angles (90 degrees) . Note: B has the largest resultant magnitude since the parallelogram has the largest diagonal. SEE RANKING QUESTION 1.|
|EXERCISES DISCUSSIONS---CHECK BACK LATER.|
|1. Fx = m*ax, where
Fx is the horizontal component of force and ax is
the horizontal component of acceleration .
(a) ax = change in velocity /change in time = (25 m/s - 0)/(0.05 s). Compute this.
(b) Find Fx = m*ax, .
(c) Find the magnitude of the force of the paper on boxer from Newton's Third Law: For every action force there is an equal and opposite reaction force.
|2. See pushboard in class quiz.. For every action force there is a reaction force. The magnitude of the action force is given and must be the same as the magnitude of the reaction force. The reaction force causes the skateboarder to accelerate away from the wall. Find the acceleration using Fx = m*ax .|
|3. READ THE SECTION ON VECTORS ! The magnitude of the speed of drops impacting your face is found as in figure 5.24. Use speed2 = 32 + 42. Take the square root of both sides to get the speed.|
|4. See previous problem. Resultant2 = 32 + 42 . Take the square root of both sides to get the magnitude of the resultant. Then divide the resultant by the mass to get the acceleration magnitude.|
|5. READ THE SECTION ON VECTORS ! Use speed2 = 1002 + 1002 . Take the square root of both sides to get the speed relative to the ground. SEE FIGURE 5.23, 5.34 AND 5.27 AS WELL AS OTHER FIGURES IN THE SECTION ON VECTO5RS.|
|6. READ THE SECTION ON VECTORS ! (a) speed2 = 32 + 42 (b) You must aim your boat at a positive angle (counterclockwise) with the rightward horizontal of the page in order to directly get across river. In other words, your aim velocity must have a component of 3 km/h upward in the diagram in the OPPOSITE direction to the downward water velocity.|