Rolling Motion: • A motion that is a combination of rotational
... H is perpendicular to r and H direction is defined by the right hand rule • unit: kg ⋅ m 2 / s • A particle does not have to travel in a circle to have angular momentum. A particle traveling in a straight line has angular momentum relative to a particular point. • If a point that the angular momentu ...
... H is perpendicular to r and H direction is defined by the right hand rule • unit: kg ⋅ m 2 / s • A particle does not have to travel in a circle to have angular momentum. A particle traveling in a straight line has angular momentum relative to a particular point. • If a point that the angular momentu ...
CONForces
... any change in its motion ◦ This means…an object at rest will stay at rest; an object in motion will stay in motion in a STRAIGHT LINE unless acted upon by an outside force. ◦ Simply put…objects like to keep doing what ...
... any change in its motion ◦ This means…an object at rest will stay at rest; an object in motion will stay in motion in a STRAIGHT LINE unless acted upon by an outside force. ◦ Simply put…objects like to keep doing what ...
ALL PHYSICS REVIEW SHEET NAME: 1. Change .0005 m to milli
... 25. If a truck 5000kg moving 45m/s collides head on with a car 3444kg moving 34m/s in the other direction, what will their speed be? 26. If a crate 344kg slides 4m down a 46º ramp at a constant speed because of a man pushing back on it, find a) force exerted by man b) work done by man on crate, c) w ...
... 25. If a truck 5000kg moving 45m/s collides head on with a car 3444kg moving 34m/s in the other direction, what will their speed be? 26. If a crate 344kg slides 4m down a 46º ramp at a constant speed because of a man pushing back on it, find a) force exerted by man b) work done by man on crate, c) w ...
WORD - hrsbstaff.ednet.ns.ca
... 9. The apparent weight (the normal force) would be largest when the elevator is accelerating upward. From the free-body diagram, with up as positive, we have FN – mg = ma. Thus FN = mg + ma. With a positive acceleration, the normal force is greater than your weight. The apparent weight would be the ...
... 9. The apparent weight (the normal force) would be largest when the elevator is accelerating upward. From the free-body diagram, with up as positive, we have FN – mg = ma. Thus FN = mg + ma. With a positive acceleration, the normal force is greater than your weight. The apparent weight would be the ...
Word
... 14) If you were to travel to the moon, you would cross a certain spot at which the net gravitational force on you (from the earth and the moon anyway) would be zero. Where would this spot be located; closer to earth, closer to the moon, or equal distance from both? Explain. Since the earth is much ...
... 14) If you were to travel to the moon, you would cross a certain spot at which the net gravitational force on you (from the earth and the moon anyway) would be zero. Where would this spot be located; closer to earth, closer to the moon, or equal distance from both? Explain. Since the earth is much ...
Name
... 10. A pumpkin with a mass of 500.0 kg sits on a level surface. You have tied a rope to the pumpkin on which you pull upward at an angle of 40.0 degrees with a force of 650.0 N. If the coefficient of friction between the pumpkin and the ground is 0.25 (a) what is the net force acting on the pumpkin? ...
... 10. A pumpkin with a mass of 500.0 kg sits on a level surface. You have tied a rope to the pumpkin on which you pull upward at an angle of 40.0 degrees with a force of 650.0 N. If the coefficient of friction between the pumpkin and the ground is 0.25 (a) what is the net force acting on the pumpkin? ...
Coefficient of Friction Worksheet
... 2. A block weighing 300 N is moved at a constant speed over a horizontal surface by a force of 50 N applied parallel to the surface. What does the “constant speed” tell you about the forces acting on the block? a. Draw a free body diagram for the block. b. What is the coefficient of kinetic friction ...
... 2. A block weighing 300 N is moved at a constant speed over a horizontal surface by a force of 50 N applied parallel to the surface. What does the “constant speed” tell you about the forces acting on the block? a. Draw a free body diagram for the block. b. What is the coefficient of kinetic friction ...
Problem: 2nd Law and Pulleys (CM-1993)
... maximum speed of 5 kilometers per hour in still water, and wish to cross a river 1 kilometer wide to a point directly across from their starting point. If the speed of the water in the river is 5 kilometers per hour, how much time is required for the crossing? (A) 0.05 hr (B) 0.1 hr (C) 1 hr (D) 10 ...
... maximum speed of 5 kilometers per hour in still water, and wish to cross a river 1 kilometer wide to a point directly across from their starting point. If the speed of the water in the river is 5 kilometers per hour, how much time is required for the crossing? (A) 0.05 hr (B) 0.1 hr (C) 1 hr (D) 10 ...
Students will understand that…
... Interpret motion graphs. Identify how acceleration, time, and velocity are related. Explain how positive and negative acceleration affect motion. Describe how to calculate the acceleration of an object. Explain how force and motion are related. Describe what inertia is and how it is rela ...
... Interpret motion graphs. Identify how acceleration, time, and velocity are related. Explain how positive and negative acceleration affect motion. Describe how to calculate the acceleration of an object. Explain how force and motion are related. Describe what inertia is and how it is rela ...
Chapter 2 Summary
... • This is the distance traveled per unit of time • Speed is a scalar quantity ...
... • This is the distance traveled per unit of time • Speed is a scalar quantity ...
Momentum
... which together provide the necessary equations to solve for the velocities after collision. For simplicity, we will confine attention to direct collisions where the bodies all travel along a straight line before and after collision. Example: Consider two balls of equal mass m one of which (ball B) i ...
... which together provide the necessary equations to solve for the velocities after collision. For simplicity, we will confine attention to direct collisions where the bodies all travel along a straight line before and after collision. Example: Consider two balls of equal mass m one of which (ball B) i ...
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.