Lecture 20: Work and Energy
... This equation states that the linear momentum vector L has a magnitude equal to (mvG) and a direction defined by vG. The angular momentum of a rigid body is defined as HG = IG Remember that the direction of HG is perpendicular to the plane of ...
... This equation states that the linear momentum vector L has a magnitude equal to (mvG) and a direction defined by vG. The angular momentum of a rigid body is defined as HG = IG Remember that the direction of HG is perpendicular to the plane of ...
Student Class ______ Date ______ MULTIPLE
... 14. This is a tricky question. Remember that the block is being pulled up the incline plane, not moving by itself. This means that we cannot say that the mechanical energy of the system is conserved. (If you like to use a formula, you could say that: W=ET, which means that work that you do, equals ...
... 14. This is a tricky question. Remember that the block is being pulled up the incline plane, not moving by itself. This means that we cannot say that the mechanical energy of the system is conserved. (If you like to use a formula, you could say that: W=ET, which means that work that you do, equals ...
1fp-lecture-notes-electronic-2015
... – KE, Torque, ang. mom., Newton’s 2nd law, rigid body rotation ...
... – KE, Torque, ang. mom., Newton’s 2nd law, rigid body rotation ...
Document
... 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 ...
Grade 5 - Detroit Public Schools
... the object is moving, it will continue to move at a constant speed in a straight line. - Two forces acting on an object in opposing directions can be of unequal strength and, therefore, are unbalanced (non-zero net force). The result will be motion (starting or speeding up) in the direction of the s ...
... the object is moving, it will continue to move at a constant speed in a straight line. - Two forces acting on an object in opposing directions can be of unequal strength and, therefore, are unbalanced (non-zero net force). The result will be motion (starting or speeding up) in the direction of the s ...
Powerpoint
... Two particles, one positively charged and one negatively charged, are held apart. Since oppositely charged objects attract one another, the particles will accelerate towards each other when released. Let W+ be the work done on the positive charge by the negative charge. Let W– be the work done on th ...
... Two particles, one positively charged and one negatively charged, are held apart. Since oppositely charged objects attract one another, the particles will accelerate towards each other when released. Let W+ be the work done on the positive charge by the negative charge. Let W– be the work done on th ...
Rotational motion is all around us
... radius is the radial distance from the axis of rotation to that point. A radius drawn from the rotation axis to any point on the body sweeps out the same angle in the same time. b. Let ri be the distance from the center of the disk to the ith particle (Figure 9-2), and let θi be the angle measured c ...
... radius is the radial distance from the axis of rotation to that point. A radius drawn from the rotation axis to any point on the body sweeps out the same angle in the same time. b. Let ri be the distance from the center of the disk to the ith particle (Figure 9-2), and let θi be the angle measured c ...
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 ...
