Work and Energy
... familiar to those who play pool. In the second case, the masses are the same. The third case shows a large mass hitting a small mass. Try the applet at this web site for experimentation. Note they use an air track so friction can be ignored. http://www.msu.edu/user/brechtjo/physics/airTrack/airTrack ...
... familiar to those who play pool. In the second case, the masses are the same. The third case shows a large mass hitting a small mass. Try the applet at this web site for experimentation. Note they use an air track so friction can be ignored. http://www.msu.edu/user/brechtjo/physics/airTrack/airTrack ...
Newton`s Second Law of Motion
... must push the car to get it started. Why can’t you push the car by remaining comfortably inside pushing against the dash? A: Pushing again the dash creates a internal force to the system, in order to accelerate the car you must have an external force to the system ...
... must push the car to get it started. Why can’t you push the car by remaining comfortably inside pushing against the dash? A: Pushing again the dash creates a internal force to the system, in order to accelerate the car you must have an external force to the system ...
dynamics intro power..
... If an apple is sitting on Mrs. Evans’ desk, it will remain there until the desk is removed (so gravity acts on it) or someone lifts it up (force). If a car is driving along a straight road at 100km/h, it will continue to do so (given the car still has gas!) until the brakes are applied (force), ther ...
... If an apple is sitting on Mrs. Evans’ desk, it will remain there until the desk is removed (so gravity acts on it) or someone lifts it up (force). If a car is driving along a straight road at 100km/h, it will continue to do so (given the car still has gas!) until the brakes are applied (force), ther ...
3.3 Projectile Motion
... The direction of the centripetal force always points toward the center of the circle and continually changes direction as the object moves. ...
... The direction of the centripetal force always points toward the center of the circle and continually changes direction as the object moves. ...
Name of Model
... axis perpendicular to hill 5. A rock sits on a hill. Draw a force diagram for the rock that allows you to determine how much of the axis parallel rock’s weight is parallel to the hill, (that would tend to to hill make it slide down the hill) and how much of the rock’s weight is perpendicular to the ...
... axis perpendicular to hill 5. A rock sits on a hill. Draw a force diagram for the rock that allows you to determine how much of the axis parallel rock’s weight is parallel to the hill, (that would tend to to hill make it slide down the hill) and how much of the rock’s weight is perpendicular to the ...
x - WordPress.com
... Circular Motion and SHM A ball rotates counterclockwise in a circle of radius 3.00 m with a constant angular speed of 8.00 rad/s. At t=0, its shadow has an x coordinate of 2.00 m and is moving to the right. Determine the x coordinate of the shadow as a function of time in SI units. Find the x ...
... Circular Motion and SHM A ball rotates counterclockwise in a circle of radius 3.00 m with a constant angular speed of 8.00 rad/s. At t=0, its shadow has an x coordinate of 2.00 m and is moving to the right. Determine the x coordinate of the shadow as a function of time in SI units. Find the x ...
Tuesday, June 6, 2006
... Galileo’s statement on natural states of matter: Any velocity once imparted to a moving body will be rigidly maintained as long as the external causes of retardation are removed!! Galileo’s statement is formulated by Newton into the 1st law of motion (Law of Inertia): In the absence of external forc ...
... Galileo’s statement on natural states of matter: Any velocity once imparted to a moving body will be rigidly maintained as long as the external causes of retardation are removed!! Galileo’s statement is formulated by Newton into the 1st law of motion (Law of Inertia): In the absence of external forc ...
AP® Physics C – Mechanics
... introduce each topic to the students and provide them with enough background information. They must then take this information and apply it to given problems or discussion questions. The problems will consist of textbook problems and old AP® Physics exam questions (used as a review before the exam). ...
... introduce each topic to the students and provide them with enough background information. They must then take this information and apply it to given problems or discussion questions. The problems will consist of textbook problems and old AP® Physics exam questions (used as a review before the exam). ...
Chapter 3 Problem Set
... F = m X a (force = mass X acceleration; here the acceleration is that generated by gravity g) W = F X d (work = force X distance), P = W/t (power = work/time) We are given the athlete’s mass (70-kg), and the distance involved is the height that he runs to (370 m) and we know that the acceleration du ...
... F = m X a (force = mass X acceleration; here the acceleration is that generated by gravity g) W = F X d (work = force X distance), P = W/t (power = work/time) We are given the athlete’s mass (70-kg), and the distance involved is the height that he runs to (370 m) and we know that the acceleration du ...
Lecture 20
... That is, momentum is conserved when the net force acting on an object is zero. This applies also to an isolated system of two of more objects (no external forces) that may be in contact - the total momentum is conserved. Compare Newton’s first law: velocity is constant when the net force is zero. ...
... That is, momentum is conserved when the net force acting on an object is zero. This applies also to an isolated system of two of more objects (no external forces) that may be in contact - the total momentum is conserved. Compare Newton’s first law: velocity is constant when the net force is zero. ...
Rotational Motion
... Solution The pivot point is at the hinges of the door, opposite to where you were pushing the door. The force you used was 50N, at a distance 1.0m from the pivot point. You hit the door perpendicular to its plane, so the angle between the door and the direction of force was 90 degrees. Since = r x ...
... Solution The pivot point is at the hinges of the door, opposite to where you were pushing the door. The force you used was 50N, at a distance 1.0m from the pivot point. You hit the door perpendicular to its plane, so the angle between the door and the direction of force was 90 degrees. Since = r x ...
Rotational Motion
... of one side. On what side of the bread will the toast land if it falls from a table 0.5 m high? If it falls from a table 1.0 m high? Assume l = 0.10 m, and ignore air resistance. When a turntable rotating at 33 rev/min is turned off, it comes to rest in 26 s. Assuming constant angular acceleration ...
... of one side. On what side of the bread will the toast land if it falls from a table 0.5 m high? If it falls from a table 1.0 m high? Assume l = 0.10 m, and ignore air resistance. When a turntable rotating at 33 rev/min is turned off, it comes to rest in 26 s. Assuming constant angular acceleration ...
Problem 1
... weight and the curve representing the force on the scale. For calculation purposes, this area can be calculated numerically by a computer. From an eyeball examination of the graph in Figure 3.2, the elevator is not moving at t1 = 2.75 s . We are given (via a numerical integration using LabView) that ...
... weight and the curve representing the force on the scale. For calculation purposes, this area can be calculated numerically by a computer. From an eyeball examination of the graph in Figure 3.2, the elevator is not moving at t1 = 2.75 s . We are given (via a numerical integration using LabView) that ...
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.