
Newton`s Second Law
... • Engine Force (FE) – Force applied to propel the train along the tracks. • Opposition Force (Fo) – friction between the tracks, wind resistance, etc. that attempts to slow the train down. • Which force was larger? • What is the acceleration of the train? Negative, Zero, or Positive. • Is this an eq ...
... • Engine Force (FE) – Force applied to propel the train along the tracks. • Opposition Force (Fo) – friction between the tracks, wind resistance, etc. that attempts to slow the train down. • Which force was larger? • What is the acceleration of the train? Negative, Zero, or Positive. • Is this an eq ...
conceptual physics c#39AC39
... Ans. No force was applied to you. The car had a centripetal force applied to it which caused the car to slide out from under you. Ask about this in class please. Extra In what direction should a force be applied to produce maximum torque? Ans. At right angles to a line that radiates out from the cen ...
... Ans. No force was applied to you. The car had a centripetal force applied to it which caused the car to slide out from under you. Ask about this in class please. Extra In what direction should a force be applied to produce maximum torque? Ans. At right angles to a line that radiates out from the cen ...
Physics 11 Dynamics - hrsbstaff.ednet.ns.ca
... standing in the wagon, the force of friction between the child and the bottom of the wagon will produce an acceleration of the feet, pulling the feet out from under the child, also making the child fall backwards. ...
... standing in the wagon, the force of friction between the child and the bottom of the wagon will produce an acceleration of the feet, pulling the feet out from under the child, also making the child fall backwards. ...
WORK AND ENERGY
... a. What potential energy will it have at the top? b. What velocity will it have at the bottom? c. What velocity will it have half way down? ...
... a. What potential energy will it have at the top? b. What velocity will it have at the bottom? c. What velocity will it have half way down? ...
Microsoft Word - Phy.. - hrsbstaff.ednet.ns.ca
... standing in the wagon, the force of friction between the child and the bottom of the wagon will produce an acceleration of the feet, pulling the feet out from under the child, also making the child fall backwards. ...
... standing in the wagon, the force of friction between the child and the bottom of the wagon will produce an acceleration of the feet, pulling the feet out from under the child, also making the child fall backwards. ...
ch15
... Resonance will occur in the forced oscillation if the natural angular frequency, , is equal to d. This is the condition when the velocity amplitude is the largest, and to some extent, also when the displacement amplitude is the largest. The adjoining figure plots displacement amplitude as a functi ...
... Resonance will occur in the forced oscillation if the natural angular frequency, , is equal to d. This is the condition when the velocity amplitude is the largest, and to some extent, also when the displacement amplitude is the largest. The adjoining figure plots displacement amplitude as a functi ...
1 - vnhsteachers
... (6) KR = ¼ (1000 kg)(0.5 m)2(314 rad/s)2 (7) KR = 6.162 x 106 J (8) P = KR / t (9) tP = KR (10) t = KR / P (11) t = 6.162 x 106 J / 1.0 x 104 watts (12) t = 616 s CONSERVATION OF MECHANICAL ENERGY The conservation of mechanical energy can be applied to rotational systems: E = K + KR STATIC EQUILIBRI ...
... (6) KR = ¼ (1000 kg)(0.5 m)2(314 rad/s)2 (7) KR = 6.162 x 106 J (8) P = KR / t (9) tP = KR (10) t = KR / P (11) t = 6.162 x 106 J / 1.0 x 104 watts (12) t = 616 s CONSERVATION OF MECHANICAL ENERGY The conservation of mechanical energy can be applied to rotational systems: E = K + KR STATIC EQUILIBRI ...
Chapter 10 PowerPoint
... The momentum of a moving object can be determined by multiplying the object’s mass and velocity. Momentum = mass x velocity ...
... The momentum of a moving object can be determined by multiplying the object’s mass and velocity. Momentum = mass x velocity ...
Lecture Notes
... Problem 24. A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 10.0 m: (a) the initially stationary spelunker is accelerated to a speed of 5.00 m/s; (b) he ...
... Problem 24. A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 10.0 m: (a) the initially stationary spelunker is accelerated to a speed of 5.00 m/s; (b) he ...
Bellringer
... If you are on a plane traveling at 400 mph, what is your speed relative to the plane? Relative to the ground? A. 0 mph B. 400 mph If someone gets up and walks to the front of the plane at 8 mph what is their speed relative to the plane? Relative to an observer on the ground? A. 8 mph B. 408 mph ...
... If you are on a plane traveling at 400 mph, what is your speed relative to the plane? Relative to the ground? A. 0 mph B. 400 mph If someone gets up and walks to the front of the plane at 8 mph what is their speed relative to the plane? Relative to an observer on the ground? A. 8 mph B. 408 mph ...
Practice Test 2
... A stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km. At the bottom of the dive, the pilot has a speed of 0.20 km/s which at that instant is not changing. What force does the plane exert on the pilot? a. b. c. d. e. ...
... A stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km. At the bottom of the dive, the pilot has a speed of 0.20 km/s which at that instant is not changing. What force does the plane exert on the pilot? a. b. c. d. e. ...
4.1 The Concepts of Force and Mass
... Two particles each have mass and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center. ...
... Two particles each have mass and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center. ...
File
... Hence moment of inertia of a body about a given the axis is numerically equal to torque acting on the body rotating with unit angular acceleration about it. We may rewrite equation (9) in vector form as τ =Iα This equation is called Fundamental equation of rotation or law of rotation.This correspon ...
... Hence moment of inertia of a body about a given the axis is numerically equal to torque acting on the body rotating with unit angular acceleration about it. We may rewrite equation (9) in vector form as τ =Iα This equation is called Fundamental equation of rotation or law of rotation.This correspon ...
PHYS2330 Intermediate Mechanics Quiz 14 Sept 2009
... This is a closed book quiz! Write the best choice in the space next to the question. 1. Given two relativistic four-momenta p1 and p2 , which of the following will have the same value in any reference frame? A. p1 p2 B. p1 · p2 C. p1 + p2 D. p1 − p2 E. p1 + p2 + p1 · p2 2. The principle moments of i ...
... This is a closed book quiz! Write the best choice in the space next to the question. 1. Given two relativistic four-momenta p1 and p2 , which of the following will have the same value in any reference frame? A. p1 p2 B. p1 · p2 C. p1 + p2 D. p1 − p2 E. p1 + p2 + p1 · p2 2. The principle moments of i ...
Gravity and Orbits
... Global Conservation Laws: •Consider a galaxy/structure which is not interacting with other galaxies/structures: •Often a good approximation •The following must be conserved: •Total energy, Total momentum, Total angular momentum P mi v i Total momentum describes overall motion i •We can eliminat ...
... Global Conservation Laws: •Consider a galaxy/structure which is not interacting with other galaxies/structures: •Often a good approximation •The following must be conserved: •Total energy, Total momentum, Total angular momentum P mi v i Total momentum describes overall motion i •We can eliminat ...
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.