Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Hunting oscillation wikipedia , lookup
Classical mechanics wikipedia , lookup
Centrifugal force wikipedia , lookup
Fictitious force wikipedia , lookup
Classical central-force problem wikipedia , lookup
Mass versus weight wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Fundamental interaction wikipedia , lookup
Multiple Choice: 1) 2) 3) 4) 5) 6) 7) 8) C D C E E C A D Part B 1. No, the speed does not change. The force is perpendicular to the velocity and so only the direction of the velocity changes, not the magnitude. The force is centripetal. 2. While it is true that the acceleration due to gravity is dependent upon distance from the center of the earth, and so does theoretically change when the height of the object changes, the earth is so large compared to the additional height that we can throw an object that this additional distance makes absolutely no detectable difference in the acceleration due to gravity. 3. The forces are similar in that both are inversely dependent upon the distance between the two objects squared. That is, they are “r-squared” forces. They both also act along a line connecting the centers of the objects. Both forces have the same mathematic form. Both are conservative forces and so have a potential energy associated with them. Both depend on (quantity 1) time (same quantity for object 2). The forces are different in that the electrostatic force can be either repulsive or attractive, while the gravitational force is always attractive. Any two objects interact gravitationally, while only objects with charge interact electrostatically. The constants are different for the two forces. Gravitational force depends on the masses of the objects, electrostatic forces depend on the charges of the objects. Part C #1 a) U=qV=(-1.6x10-19)(10)=-1.6x10-18 J b) U=qV=(-1.6x10-19)(20-10)=-1.6x10-18 J c)-U=KE so +1.6x10-18=1/2mv2 v = 1.87 x106 m/s d) FB=qvxB=qvB=(1.6x10-19)(v)(5x10-4) =1.50 x 10-16 N #2 a) F2net F1,2 F2,3 q q qq k 1 2 k 3 2 r12 2 r32 2 q k 2 (q1 q3 ) 16 (8.99x109 )(2 x10 6 ) (3x10 6 ) 16 3.37x103 N B) Left- no partial credit (+3 points) C) E y ,net E1, y E2, y E3, y q q q k 1 Sin45o k 2 k 3 Sin45o r12 r22 r32 k Cos45 q (q1 q3 ) k 2 32 16 (8.99x109 )(.707)(7 x106 ) (8.99x109 )(2 x106 ) 32 16 266 N/C D) E x, net E1, x E2, x E3, x q q k 1 Cos45o 0 k 3 Cos45o r12 r32 k Cos45 (q1 q3 ) 32 (8.99x109 )(.707)(3x106 ) 32 596 N/C E) Up and to the right