Document
... The equations above show that the center of mass of a system of particles moves as though all the system's mass were concentrated there, and that the vector sum of all the external forces were applied there. A dramatic example is given in the figure. In a fireworks display a rocket is launched and m ...
... The equations above show that the center of mass of a system of particles moves as though all the system's mass were concentrated there, and that the vector sum of all the external forces were applied there. A dramatic example is given in the figure. In a fireworks display a rocket is launched and m ...
Balanced And Unbalanced Forces We perform different types of
... The toy car comes to rest after some time because of the frictional force between the moving wheels of the car and the rough floor. This force acts in the direction opposite to the direction of motion of the car. This means that an unbalanced force acts on the car in the direction opposite to the di ...
... The toy car comes to rest after some time because of the frictional force between the moving wheels of the car and the rough floor. This force acts in the direction opposite to the direction of motion of the car. This means that an unbalanced force acts on the car in the direction opposite to the di ...
Center of Mass
... • If this body is executing motion under the effect of some external forces acting on it then it has been found that there is a point in the system , where if whole mass of the system is supposed to be concentrated and the nature the motion executed by the system remains unaltered when force acting ...
... • If this body is executing motion under the effect of some external forces acting on it then it has been found that there is a point in the system , where if whole mass of the system is supposed to be concentrated and the nature the motion executed by the system remains unaltered when force acting ...
quiz and answers ch4 sec 1-2
... 2. The free-body diagram shown above represents a car being pulled by a towing cable. In the diagram, which of the following is the gravitational force acting on the car? a. 5800 N c. 14 700 N b. 775 N d. 13 690 N 3. The free-body diagram shown above represents a car being pulled by a towing cable. ...
... 2. The free-body diagram shown above represents a car being pulled by a towing cable. In the diagram, which of the following is the gravitational force acting on the car? a. 5800 N c. 14 700 N b. 775 N d. 13 690 N 3. The free-body diagram shown above represents a car being pulled by a towing cable. ...
Lecture 1
... A force accelerates an object and the acceleration is proportional to the force: F=m*a The constant m is the mass of the object Force ...
... A force accelerates an object and the acceleration is proportional to the force: F=m*a The constant m is the mass of the object Force ...
8th Grade Physical Science
... Energy is constantly changing from one form to another. As energy changes from one form to another, it is never created or destroyed. Many times it takes a whole series of energy conversions to do a certain job. For example, just to get the energy to make a piece of toast, there are several ...
... Energy is constantly changing from one form to another. As energy changes from one form to another, it is never created or destroyed. Many times it takes a whole series of energy conversions to do a certain job. For example, just to get the energy to make a piece of toast, there are several ...
Solutions to Assignment #1
... #4. A long jumper takes off with a velocity of 11.00 m/s at an angle of 25.0o. The athlete’s centre of gravity was 1.250 m above the ground at takeoff and 0.500 m above the ground at landing. (a) Compute the rectangular components of the takeoff velocity. (b) What will be the flight time? (c) How fa ...
... #4. A long jumper takes off with a velocity of 11.00 m/s at an angle of 25.0o. The athlete’s centre of gravity was 1.250 m above the ground at takeoff and 0.500 m above the ground at landing. (a) Compute the rectangular components of the takeoff velocity. (b) What will be the flight time? (c) How fa ...
Rigid Body Simulation (1)
... where G = 6.67x10-11, a universal constant D = distance between centers of mass ...
... where G = 6.67x10-11, a universal constant D = distance between centers of mass ...
Study Guide for GLO Conceptual Physics
... Speed (a scalar quantity) is the rate at which a moving object covers distance in (m/s) To have a scalar quantity, you do not have direction (distance, time) SP1.c Compare graphically and algebraically the relationships among position, velocity, acceleration, and time. Topics: Position, velocity, ...
... Speed (a scalar quantity) is the rate at which a moving object covers distance in (m/s) To have a scalar quantity, you do not have direction (distance, time) SP1.c Compare graphically and algebraically the relationships among position, velocity, acceleration, and time. Topics: Position, velocity, ...
Wednesday, Oct. 29, 2008
... momentum and the kinetic energy m1 v12i v12f m2 v22i v22 f are conserved. Therefore, the final speeds in an elastic collision m1 v1i v1 f v1i v1 f m2 v2i v2 f v2i v2 f can be obtained in terms of initial From momentum m1 v1i v1 f m2 v2i v2 f speeds as cons ...
... momentum and the kinetic energy m1 v12i v12f m2 v22i v22 f are conserved. Therefore, the final speeds in an elastic collision m1 v1i v1 f v1i v1 f m2 v2i v2 f v2i v2 f can be obtained in terms of initial From momentum m1 v1i v1 f m2 v2i v2 f speeds as cons ...
Gravitation and Momentum
... • If a car hits a haystack or the same car hits a wall, momentum is decreased by same impulse – the same products of force and time. • However, impact force is greater into the wall than it is into the haystack as the haystack extends impact time, lessening the impact force. • Impact time is the tim ...
... • If a car hits a haystack or the same car hits a wall, momentum is decreased by same impulse – the same products of force and time. • However, impact force is greater into the wall than it is into the haystack as the haystack extends impact time, lessening the impact force. • Impact time is the tim ...
Rotational Motion
... A rotating object is one that spins on a fixed axis. The position and direction of the rotation axis will remain constant. The position of some part of the object can be specified with standard cartesian coordinates, (x,y). All objects will be assumed to rotate in a circular path of constant radius. ...
... A rotating object is one that spins on a fixed axis. The position and direction of the rotation axis will remain constant. The position of some part of the object can be specified with standard cartesian coordinates, (x,y). All objects will be assumed to rotate in a circular path of constant radius. ...
Practice test (Chapters 10
... The rigid body shown rotates about an axis through its center of mass and perpendicular to the paper. If M = 2.0 kg and L = 80 cm, what is the kinetic energy of this object when its angular speed about this axis is equal to 5.0 rad/s? Neglect the mass of the connecting rod and treat the masses as pa ...
... The rigid body shown rotates about an axis through its center of mass and perpendicular to the paper. If M = 2.0 kg and L = 80 cm, what is the kinetic energy of this object when its angular speed about this axis is equal to 5.0 rad/s? Neglect the mass of the connecting rod and treat the masses as pa ...
Chapter 6 Work and Energy
... and over the displacement s, the speed of the car will increase. Newton's 2nd law: acceleration of the car, a = F mCar Starting with velocity v0 , find the final speed. ...
... and over the displacement s, the speed of the car will increase. Newton's 2nd law: acceleration of the car, a = F mCar Starting with velocity v0 , find the final speed. ...
wbm-physics
... The work done by conservative forces: Can always be expressed as the difference between the initial and final values of a potential energy function. Is reversible Is path-independent (It only depends on the starting and ending points.) Equals zero when the starting and ending points are the sa ...
... The work done by conservative forces: Can always be expressed as the difference between the initial and final values of a potential energy function. Is reversible Is path-independent (It only depends on the starting and ending points.) Equals zero when the starting and ending points are the sa ...
Part 2 - Haiku
... the amount of applied force, Fapplied, necessary to overcome static friction, Fstatic and at this moment, they are nearly equal forces, so assume Fapplied = Fstatic . Record the minimum applied force as Fstatic. 3. While the crate is in motion, let go of the slider, observe the change in the force v ...
... the amount of applied force, Fapplied, necessary to overcome static friction, Fstatic and at this moment, they are nearly equal forces, so assume Fapplied = Fstatic . Record the minimum applied force as Fstatic. 3. While the crate is in motion, let go of the slider, observe the change in the force v ...
Newton`s Laws of Motion: PowerPoint
... – is equal to the mass of an object (m) times the acceleration due to ...
... – is equal to the mass of an object (m) times the acceleration due to ...
Fall 2008 - BYU Physics and Astronomy
... the spring slowly stretch out so that the mass eventually hangs without moving at a distance of 20 cm below where it started. What is the spring constant k? a. less than 10 N/m b. 10 - 19 c. 19 - 28 d. 28 - 37 e. 37 - 46 f. 46 - 55 g. more than 55 N/m Problem 35. You want to measure the spring const ...
... the spring slowly stretch out so that the mass eventually hangs without moving at a distance of 20 cm below where it started. What is the spring constant k? a. less than 10 N/m b. 10 - 19 c. 19 - 28 d. 28 - 37 e. 37 - 46 f. 46 - 55 g. more than 55 N/m Problem 35. You want to measure the spring const ...
Classical central-force problem
In classical mechanics, the central-force problem is to determine the motion of a particle under the influence of a single central force. A central force is a force that points from the particle directly towards (or directly away from) a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center. In many important cases, the problem can be solved analytically, i.e., in terms of well-studied functions such as trigonometric functions.The solution of this problem is important to classical physics, since many naturally occurring forces are central. Examples include gravity and electromagnetism as described by Newton's law of universal gravitation and Coulomb's law, respectively. The problem is also important because some more complicated problems in classical physics (such as the two-body problem with forces along the line connecting the two bodies) can be reduced to a central-force problem. Finally, the solution to the central-force problem often makes a good initial approximation of the true motion, as in calculating the motion of the planets in the Solar System.