2002 - University of Utah Physics
... Please note that there is a separate booklet for each numbered question (i.e., use booklet #1 for problem #1, etc.). To receive full credit, not only should the correct solutions be given, but a sufficient number of steps should be given so that a faculty grader can follow your reasoning. Define all ...
... Please note that there is a separate booklet for each numbered question (i.e., use booklet #1 for problem #1, etc.). To receive full credit, not only should the correct solutions be given, but a sufficient number of steps should be given so that a faculty grader can follow your reasoning. Define all ...
Centripetal acceleration
... if the units of each term of an equation match, the formula is not necessarily correct (some unitless factors might be missing), but if the units of each term are not the same then you can be sure the formula is incorrect. OK, that's good, but it would be better to have an intuitive feel for why the ...
... if the units of each term of an equation match, the formula is not necessarily correct (some unitless factors might be missing), but if the units of each term are not the same then you can be sure the formula is incorrect. OK, that's good, but it would be better to have an intuitive feel for why the ...
shm-intro - Mrs Physics
... position again? 0.29 s 9. It takes a force of 60 N to compress the spring of a popgun 0.10 m to load a 0.200 kg ball. With what speed will the ball leave the gun? 10. How much would a spring scale with k=120 N/m stretch, if it had 3.75 J of work done on it? 11. A block of mass 0.50 kg is placed on a ...
... position again? 0.29 s 9. It takes a force of 60 N to compress the spring of a popgun 0.10 m to load a 0.200 kg ball. With what speed will the ball leave the gun? 10. How much would a spring scale with k=120 N/m stretch, if it had 3.75 J of work done on it? 11. A block of mass 0.50 kg is placed on a ...
ap physics b - New Jersey Center for Teaching and Learning
... a. Decomposition of vectors b. Composition of vectors c. Review - Motion in one dimension d. Motion in two dimensions (Independence of perpendicular components) e. Displacement in two dimensions f. Velocity in two dimensions g. Acceleration in two dimensions h. Projectile motion i. Review - Uniform ...
... a. Decomposition of vectors b. Composition of vectors c. Review - Motion in one dimension d. Motion in two dimensions (Independence of perpendicular components) e. Displacement in two dimensions f. Velocity in two dimensions g. Acceleration in two dimensions h. Projectile motion i. Review - Uniform ...
Dynamics Rewrite Problems 1. A 0.40 kg toy car moves at constant
... 12. The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.25. The weight of the object is 20N. What is the force of friction? 13. The force of friction between an object and the surface upon which it is sliding is 15 N. The mass of the object is 20.0 kg. ...
... 12. The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.25. The weight of the object is 20N. What is the force of friction? 13. The force of friction between an object and the surface upon which it is sliding is 15 N. The mass of the object is 20.0 kg. ...
Solutions to Practice Problems for Test 2
... trial solution Y2 = F sin t + G cos t. For (DE1) and (DE2) we find the particular solutions yp1 , yp2 , so the solution to our problem will be y = yc + yp1 + yp2 . (b) y 00 + 2y 0 + 5y = 4e−t cos 2t Solution The complementary solution is yc = e−t (A cos 2t + B sin 2t). Take the trial solution Y = te ...
... trial solution Y2 = F sin t + G cos t. For (DE1) and (DE2) we find the particular solutions yp1 , yp2 , so the solution to our problem will be y = yc + yp1 + yp2 . (b) y 00 + 2y 0 + 5y = 4e−t cos 2t Solution The complementary solution is yc = e−t (A cos 2t + B sin 2t). Take the trial solution Y = te ...
2005 - State Examination Commission
... (c) Calculate the potential energy of an object of mass 0.5 kg at a height of 100 m above the surface of the earth. (g = 9.8 m s–2) (d) Give an example of (i) a transverse wave, (ii) a longitudinal wave. (e) A ray of light enters a 45º right-angled glass prism as shown in Fig. 1. Copy the diagram an ...
... (c) Calculate the potential energy of an object of mass 0.5 kg at a height of 100 m above the surface of the earth. (g = 9.8 m s–2) (d) Give an example of (i) a transverse wave, (ii) a longitudinal wave. (e) A ray of light enters a 45º right-angled glass prism as shown in Fig. 1. Copy the diagram an ...
Systems of Particles
... Example of elastic collision A small ball of mass m is aligned above a larger ball of mass M=0.63 kg (with a slight separation, as with the baseball and basketball of Fig. 9-70a), and the two are dropped simultaneously from a height of h=1.8m. (Assume the radius of each ball is negligible relative ...
... Example of elastic collision A small ball of mass m is aligned above a larger ball of mass M=0.63 kg (with a slight separation, as with the baseball and basketball of Fig. 9-70a), and the two are dropped simultaneously from a height of h=1.8m. (Assume the radius of each ball is negligible relative ...
Physics Fall Exam Study Guide
... If you throw a ball vertically into the air 20 m and catch it at the same height you released it: What was the speed at the top of the path? _________________________________________ Compare the velocity when you let it go and when you catch it: _________________________ Compare the velocities ...
... If you throw a ball vertically into the air 20 m and catch it at the same height you released it: What was the speed at the top of the path? _________________________________________ Compare the velocity when you let it go and when you catch it: _________________________ Compare the velocities ...
Chapter 3: Motion and Forces Goals of Period 3
... right or the left side). In the case of the equation s = D/t, we must cancel t from the right side of the equation and move t to the left side. We need an operation that allows us to cancel t. Multiplying both sides of the equation by t allows cancellation of t from the right side. st ...
... right or the left side). In the case of the equation s = D/t, we must cancel t from the right side of the equation and move t to the left side. We need an operation that allows us to cancel t. Multiplying both sides of the equation by t allows cancellation of t from the right side. st ...
