Final Exam Practice questions
... 19) The velocity of an airplane with respect to the ground is 200 m/s at an angle of 30 degrees NORTH of EAST. The velocity of the airplane with respect to the air is 150 m/s at an angle of 60 degrees NORTH of EAST. What is the velocity of the air with respect to the ground? a) 158 m/s at 16.9 degr ...
... 19) The velocity of an airplane with respect to the ground is 200 m/s at an angle of 30 degrees NORTH of EAST. The velocity of the airplane with respect to the air is 150 m/s at an angle of 60 degrees NORTH of EAST. What is the velocity of the air with respect to the ground? a) 158 m/s at 16.9 degr ...
Physics 207: Lecture 2 Notes
... The constant K is called the “coefficient of kinetic friction”. Physics 207: Lecture 6, Pg 40 ...
... The constant K is called the “coefficient of kinetic friction”. Physics 207: Lecture 6, Pg 40 ...
Friction
... • The negative sign for acceleration a is dropped because k is a ratio of forces that does not depend on direction. • Maximum stopping distance occurs when the tire is rotating. When this happens, a = -s·g. • Otherwise, use a = -k·g to find the acceleration, then use a velocity equation to find d ...
... • The negative sign for acceleration a is dropped because k is a ratio of forces that does not depend on direction. • Maximum stopping distance occurs when the tire is rotating. When this happens, a = -s·g. • Otherwise, use a = -k·g to find the acceleration, then use a velocity equation to find d ...
Circular Motion
... we’ve picked an unusual coordinate system. Not down the inclined plane, but aligned with the radial direction. That’s because we want to determine the component of any force or forces that may act as a centripetal force. We are ignoring friction so the only two forces to consider are the weight mg a ...
... we’ve picked an unusual coordinate system. Not down the inclined plane, but aligned with the radial direction. That’s because we want to determine the component of any force or forces that may act as a centripetal force. We are ignoring friction so the only two forces to consider are the weight mg a ...
Lesson 1 - SchoolRack
... change of motion Newton’s first law of motion states that an object will remain at rest or in constant straight-line motion unless unbalanced forces act on the object. • Newton’s second law of motion states that the acceleration of an object increases as the force acting on it increases and decrease ...
... change of motion Newton’s first law of motion states that an object will remain at rest or in constant straight-line motion unless unbalanced forces act on the object. • Newton’s second law of motion states that the acceleration of an object increases as the force acting on it increases and decrease ...
Dynamics Chapter Problems
... surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses. Include al ...
... surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses. Include al ...
Physics - Oak Park Unified School District
... special case: if vo = 0, then vt = 2vav b. falling objects 1. all objects fall with the same constant acceleration. (Galileo) 2. g, at sea level is about 9.80 m/s2 5. positive and negative case a. direction can be "forward (+) or backward (–) b. velocity and displacement always have same sign c. acc ...
... special case: if vo = 0, then vt = 2vav b. falling objects 1. all objects fall with the same constant acceleration. (Galileo) 2. g, at sea level is about 9.80 m/s2 5. positive and negative case a. direction can be "forward (+) or backward (–) b. velocity and displacement always have same sign c. acc ...
Part III: Movement Analysis – Learning Outcomes
... turn about its axis of rotation with constant angular momentum unless an external force is exerted upon it. The angular form of Newton’s second law - The angular acceleration of a body is proportional to the torque causing it and takes place in the direction in which the torque acts. The angular for ...
... turn about its axis of rotation with constant angular momentum unless an external force is exerted upon it. The angular form of Newton’s second law - The angular acceleration of a body is proportional to the torque causing it and takes place in the direction in which the torque acts. The angular for ...
Rotational dynamics
... A turntable of mass 5kg and radius 25 cm is rotating at 10 radsˉ¹. A metal ring of mass 2 kg and radius 10 cm is dropped over the centre of the turntable. a) Find the new angular velocity of the system. b) Using rotational energy determine whether this is an elastic or an ...
... A turntable of mass 5kg and radius 25 cm is rotating at 10 radsˉ¹. A metal ring of mass 2 kg and radius 10 cm is dropped over the centre of the turntable. a) Find the new angular velocity of the system. b) Using rotational energy determine whether this is an elastic or an ...
Rotational Motion - My Teacher Pages
... the Lever Arm • Torque can also be calculated using the concept of a lever arm • Lever arm is the perpendicular distance from the axis of rotation to the line containing the force vector • “Line of action” ...
... the Lever Arm • Torque can also be calculated using the concept of a lever arm • Lever arm is the perpendicular distance from the axis of rotation to the line containing the force vector • “Line of action” ...
Page 24 #10
... For the component vectors of the cannonball’s motion, the horizontal component is always the same and only the vertical component changes. At the top of the path the vertical component shrinks to zero, so the velocity there is the same as the horizontal component of velocity at all other points. Eve ...
... For the component vectors of the cannonball’s motion, the horizontal component is always the same and only the vertical component changes. At the top of the path the vertical component shrinks to zero, so the velocity there is the same as the horizontal component of velocity at all other points. Eve ...
Chapter 1 D`Alembert`s principle and applications
... 5. The cross brace breaks on a step ladder erected on a flat, icy (frictionless) sidewalk as illustrated in figure 1.4.4, causing the ladder to collapse. Fortunately, no one is standing on it at the time. Compute the governing equation for φ(t) as the ladder collapses using d’Alembert’s principle. A ...
... 5. The cross brace breaks on a step ladder erected on a flat, icy (frictionless) sidewalk as illustrated in figure 1.4.4, causing the ladder to collapse. Fortunately, no one is standing on it at the time. Compute the governing equation for φ(t) as the ladder collapses using d’Alembert’s principle. A ...
