Chapter 5 Work and Energy
... by a spring with a relaxed length of 0.115 m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m? ...
... by a spring with a relaxed length of 0.115 m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m? ...
PowerPoint Presentation - 5. Universal Laws of Motion
... acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics which are true of free-falling objects: – Free-falling objects do not encounter air resistance. – All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/ ...
... acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics which are true of free-falling objects: – Free-falling objects do not encounter air resistance. – All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/ ...
Mechanical Equilibrium
... an object is directly proportional to the magnitude of the net force, is in the same direction as the net force, and is inversely proportional to the mass of the object. Newton’s third law states that whenever one object exerts a force on a second object, the second object exerts an equal and opposi ...
... an object is directly proportional to the magnitude of the net force, is in the same direction as the net force, and is inversely proportional to the mass of the object. Newton’s third law states that whenever one object exerts a force on a second object, the second object exerts an equal and opposi ...
Quantum Mechanics Practice Problems Solutions
... in the left hand third of the box (i.e. between 0 and ⅓L)? 33%, classical mechanics would say that the particle is equally likely to be anywhere in the box so there is a 33% chance of it being in any third of the box ...
... in the left hand third of the box (i.e. between 0 and ⅓L)? 33%, classical mechanics would say that the particle is equally likely to be anywhere in the box so there is a 33% chance of it being in any third of the box ...
PHY 140Y – FOUNDATIONS OF PHYSICS 1999
... A block of mass m is on a level table top. It is pulled by means of two strings, aligned in the x and y directions, respectively. The tensions in the two strings have equal magnitudes T. The coefficient of kinetic friction between the block and the table top is µk. Under the influence of the two str ...
... A block of mass m is on a level table top. It is pulled by means of two strings, aligned in the x and y directions, respectively. The tensions in the two strings have equal magnitudes T. The coefficient of kinetic friction between the block and the table top is µk. Under the influence of the two str ...
The Force Be With You
... • Recall that the more massive an object is, the more inertia it has. • Therefore, more massive objects are harder to accelerate. • Acceleration is inversely proportional to mass. – This means that when one gets bigger, the other one gets smaller. ...
... • Recall that the more massive an object is, the more inertia it has. • Therefore, more massive objects are harder to accelerate. • Acceleration is inversely proportional to mass. – This means that when one gets bigger, the other one gets smaller. ...
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 ...
Pledged Problems 8
... – Print the phrase ‘Pledged Problems 8’ and the due date. – Print the times at which you started and finished the problems. – Write and sign the Pledge, with the understanding you may consult the resources described above. ...
... – Print the phrase ‘Pledged Problems 8’ and the due date. – Print the times at which you started and finished the problems. – Write and sign the Pledge, with the understanding you may consult the resources described above. ...
Describing Motion Verbally with Speed and Velocity
... 11. Earth-orbiting astronauts feel weightless in space because _____. Choose all that apply. a. They are in free-fall motion. b. There is an absence of contact forces acting upon their bodies. c. The weight of objects diminish to close to 0 N at these distances from Earth's center. d. There is no gr ...
... 11. Earth-orbiting astronauts feel weightless in space because _____. Choose all that apply. a. They are in free-fall motion. b. There is an absence of contact forces acting upon their bodies. c. The weight of objects diminish to close to 0 N at these distances from Earth's center. d. There is no gr ...
Net Force: a resultant force acting on object
... free-body diagram for them Choose a convenient coordinate system for each object Apply Newton’s second law. The x- and y-components of Newton second law should be taken from the vector equation and written individually. This often results in two equations and two unknowns Solve for the desired unkno ...
... free-body diagram for them Choose a convenient coordinate system for each object Apply Newton’s second law. The x- and y-components of Newton second law should be taken from the vector equation and written individually. This often results in two equations and two unknowns Solve for the desired unkno ...
Document
... • or in uniform motion in a straight line, • unless it is compelled to change by an externally imposed force. ...
... • or in uniform motion in a straight line, • unless it is compelled to change by an externally imposed force. ...
Chapter 3
... 3.2 Vectors vector – represented by an arrow whose length is the magnitude of a physical entity and whose orientation shows how the physical entity is directed. ...
... 3.2 Vectors vector – represented by an arrow whose length is the magnitude of a physical entity and whose orientation shows how the physical entity is directed. ...
Guided Practice—Student Copy
... across the room and stretch it tightly. Repeat with another fishing line making them parallel with each other. Have students blow up balloons of various shapes and sizes without tying them. They should just hold them in place until ready to race. When it is their turn, attach their balloon to the st ...
... across the room and stretch it tightly. Repeat with another fishing line making them parallel with each other. Have students blow up balloons of various shapes and sizes without tying them. They should just hold them in place until ready to race. When it is their turn, attach their balloon to the st ...
Chapter 10b
... Four small spheres are mounted on the corners of a frame as shown. a) What is the rotational energy of the system if it is rotated about the z-axis (out of page) with an angular velocity of 5 rad/s b) What is the rotational energy if the system is rotated about the yaxis? (M = 5 kg; m = 2 kg; a = 1. ...
... Four small spheres are mounted on the corners of a frame as shown. a) What is the rotational energy of the system if it is rotated about the z-axis (out of page) with an angular velocity of 5 rad/s b) What is the rotational energy if the system is rotated about the yaxis? (M = 5 kg; m = 2 kg; a = 1. ...
Slide 1
... 6) At the strike of midnight, a new-year’s reveler shoots his .357 Remington, aimed upwards at a 45° angle. How far away does the bullet land? vmuzzle = 46 m/s R=(v2/g)sin2θ = [(46 m/s)2/9.8 m/s] × (sin90°) R=216 m At what time does the bullet go through your living room window? (Magic: vhorizontal ...
... 6) At the strike of midnight, a new-year’s reveler shoots his .357 Remington, aimed upwards at a 45° angle. How far away does the bullet land? vmuzzle = 46 m/s R=(v2/g)sin2θ = [(46 m/s)2/9.8 m/s] × (sin90°) R=216 m At what time does the bullet go through your living room window? (Magic: vhorizontal ...
Lecture 18
... Note that the MG moment equation may be replaced by a moment summation about any arbitrary point. Summing the moment about the center of rotation O yields MO = IGa + rG m (aG) t = (IG + m (rG)2 ) a From the parallel axis theorem, IO = IG + m(rG)2, therefore the term in parentheses represents IO. C ...
... Note that the MG moment equation may be replaced by a moment summation about any arbitrary point. Summing the moment about the center of rotation O yields MO = IGa + rG m (aG) t = (IG + m (rG)2 ) a From the parallel axis theorem, IO = IG + m(rG)2, therefore the term in parentheses represents IO. C ...
Explaining motion
... of the object does not change Law 2 – if there is a resultant force acting on an object, the momentum will change (c.o.m.=r.f x time) and is in the same direction ...
... of the object does not change Law 2 – if there is a resultant force acting on an object, the momentum will change (c.o.m.=r.f x time) and is in the same direction ...
Study Guide Exercises
... 5. Circle the letter of each statement about force and acceleration that is true. a. Balanced forces cause constant acceleration. b. The forces acting on an object at rest are unbalanced. QA net force acting on an object causes acceleration. d. Force is not required for an object to accelerate. 6. T ...
... 5. Circle the letter of each statement about force and acceleration that is true. a. Balanced forces cause constant acceleration. b. The forces acting on an object at rest are unbalanced. QA net force acting on an object causes acceleration. d. Force is not required for an object to accelerate. 6. T ...
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.