Lab M14 – Pulleys
... a) Using a force diagram for the hanging mass and the fact that the string pulls with the same force at all places, assuming that there is no friction with the pulleys. How does Newton’s First Law apply here ? b) Using the fact that for an ideal machine, the work done by the effort force equals the ...
... a) Using a force diagram for the hanging mass and the fact that the string pulls with the same force at all places, assuming that there is no friction with the pulleys. How does Newton’s First Law apply here ? b) Using the fact that for an ideal machine, the work done by the effort force equals the ...
electric circuits
... In this section, we will learn how: Second-order differential equations are applied to the vibration of springs and electric circuits. ...
... In this section, we will learn how: Second-order differential equations are applied to the vibration of springs and electric circuits. ...
17 Energy in SHM - Blue Valley Schools
... 1. In the introduction, we claimed that the gravitational potential energy could be ignored if the displacement used in the elastic potential energy was measured from the hanging equilibrium position. First write the total mechanical energy (kinetic, gravitational potential, and elastic potential en ...
... 1. In the introduction, we claimed that the gravitational potential energy could be ignored if the displacement used in the elastic potential energy was measured from the hanging equilibrium position. First write the total mechanical energy (kinetic, gravitational potential, and elastic potential en ...
Circular Motion and Gravitation Notes 1 – Centripetal Acceleration
... A 0.50 kg mass sits on a frictionless table and is attached to hanging weight. The 0.50 kg mass is whirled in a circle of radius 0.20 m at 2.3 m/s. Calculate the centripetal force acting on the mass. ...
... A 0.50 kg mass sits on a frictionless table and is attached to hanging weight. The 0.50 kg mass is whirled in a circle of radius 0.20 m at 2.3 m/s. Calculate the centripetal force acting on the mass. ...
test1
... backward (recoils) at a speed of 1.4 m/s. Which principle would you use to solve the problem? Calculate the frog’s horizontal jumping speed and indicate the direction too. ...
... backward (recoils) at a speed of 1.4 m/s. Which principle would you use to solve the problem? Calculate the frog’s horizontal jumping speed and indicate the direction too. ...
![Multiple choice questions [60 points]](http://s1.studyres.com/store/data/002785313_1-b2734444f348f25d9b46ea15f542520b-300x300.png)
Work is a force that moves through a distance
... 5. 1N equals 1kgm/s2. What is a Dyne equal to? 6 How much force is required to accelerate a 1800kg car at 3.00m/s/s? 7. What is the weight of the 1800kg car in question G? 8. What is the difference between mass and weight? 9. What unit is weight measured in? 10. What is inertia? 11. What do automobi ...
... 5. 1N equals 1kgm/s2. What is a Dyne equal to? 6 How much force is required to accelerate a 1800kg car at 3.00m/s/s? 7. What is the weight of the 1800kg car in question G? 8. What is the difference between mass and weight? 9. What unit is weight measured in? 10. What is inertia? 11. What do automobi ...
Force Practice Problems Name: Per: ______ Answer the following
... 5. Two forces act on a 60.0 kg mass. One force is pulling the mass in the +x direction with 180.N. The other force is pulling the mass in the –x direction with 210.N. a. Calculate the NET force acting on the mass. b. Calculate the NET acceleration of the mass, including direction. c. If the two forc ...
... 5. Two forces act on a 60.0 kg mass. One force is pulling the mass in the +x direction with 180.N. The other force is pulling the mass in the –x direction with 210.N. a. Calculate the NET force acting on the mass. b. Calculate the NET acceleration of the mass, including direction. c. If the two forc ...
If a simple pendulum oscillates with an amplitude 50 mm and time
... velocity is (a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s [Option (b)]. The following question appeared in Kerala Engineering Entrance 2005 test paper: A ...
... velocity is (a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s [Option (b)]. The following question appeared in Kerala Engineering Entrance 2005 test paper: A ...