PHY 101 Lecture Notes
... Distance traveled for an object that starts at rest: d = ½ acceleration * (time)2 (general) =½g*t2 (for free fall) Distance traveled for an object that starts with an initial speed d = initial velocity * time + ½ acceleration * (time)2 = initial velocity * t - 1/2 g t2 Remember to use correct units: ...
... Distance traveled for an object that starts at rest: d = ½ acceleration * (time)2 (general) =½g*t2 (for free fall) Distance traveled for an object that starts with an initial speed d = initial velocity * time + ½ acceleration * (time)2 = initial velocity * t - 1/2 g t2 Remember to use correct units: ...
CHAPTER 1 INTRODUCTION TO PHYSICS
... If the gravitational acceleration on the moon is 1.6 ms-2, compare the time taken for the object to fall from the same height on the Moon. A Same as t s B Less than t s C Longer than t s ...
... If the gravitational acceleration on the moon is 1.6 ms-2, compare the time taken for the object to fall from the same height on the Moon. A Same as t s B Less than t s C Longer than t s ...
assignments
... which relationships to utilize, how they apply to the problem, and where to apply them. Solution: The critical point is when the particle is at the highest point of the loop to loop. The Kinetic Energy at this point is equal to the change in Potential Energy so KE = ½ m v2 = m g (h – D). In addition ...
... which relationships to utilize, how they apply to the problem, and where to apply them. Solution: The critical point is when the particle is at the highest point of the loop to loop. The Kinetic Energy at this point is equal to the change in Potential Energy so KE = ½ m v2 = m g (h – D). In addition ...
Work, Energy and Power
... • A 20kg suitcase is raised vertically 3m above a platform by a conveyor belt. How much work is done on the suitcase? • A 100 N force is applied to move a 15kg object a horizontal distance of 5m at constant speed. • A 100N force is applied at an angle of 30° to the horizontal to move a 15kg object a ...
... • A 20kg suitcase is raised vertically 3m above a platform by a conveyor belt. How much work is done on the suitcase? • A 100 N force is applied to move a 15kg object a horizontal distance of 5m at constant speed. • A 100N force is applied at an angle of 30° to the horizontal to move a 15kg object a ...
PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION
... their loadings, are considered to be symmetrical with respect to a fixed reference plane. In this case the path of motion of each particle of the body is a plane curve parallel to a fixed reference plane. Since the motion of the body may be viewed within the reference plane, all the forces (and coup ...
... their loadings, are considered to be symmetrical with respect to a fixed reference plane. In this case the path of motion of each particle of the body is a plane curve parallel to a fixed reference plane. Since the motion of the body may be viewed within the reference plane, all the forces (and coup ...
Circular Motion - Saptarshi Classes
... 36) A coin placed on a rotating table just slips when it is placed at a distance 4r from the centre, on doubling the angular velocity of the table, the coin will just slip now the coin is at a distance from centre is a) 4r b) r/4 c) r d) 2 r 37) A car is moving along a horizontal curve of radius 20 ...
... 36) A coin placed on a rotating table just slips when it is placed at a distance 4r from the centre, on doubling the angular velocity of the table, the coin will just slip now the coin is at a distance from centre is a) 4r b) r/4 c) r d) 2 r 37) A car is moving along a horizontal curve of radius 20 ...
THE NOT SO SIMPLE PENDULUM
... The deriving force for the motion of the mass is mgSin(θ). For small angles this can be written as mgx/L which shows that the force is proportional to the horizontal position x. The acceleration, gx/L is also proportional to x. The graph of acceleration versus position in Fig. 4 shows this behaviour ...
... The deriving force for the motion of the mass is mgSin(θ). For small angles this can be written as mgx/L which shows that the force is proportional to the horizontal position x. The acceleration, gx/L is also proportional to x. The graph of acceleration versus position in Fig. 4 shows this behaviour ...
Experiment: Uniform Circular Motion
... The figure below shows a free body diagram of the forces acting on Ozobot while in uniform circular motion. Since Ozobot remains on its level surface and does not move upward, the normal force and the force of gravity are equal in magnitude and oppositely directed. The net force is therefore the for ...
... The figure below shows a free body diagram of the forces acting on Ozobot while in uniform circular motion. Since Ozobot remains on its level surface and does not move upward, the normal force and the force of gravity are equal in magnitude and oppositely directed. The net force is therefore the for ...
Chapter 11 Force and Newton`s Laws
... Texture – A rougher surface = greater friction. A smoother surface = less friction. Mass/Weight – The greater the mass or weight of an object, the more friction it will create. Fluids – Fluids reduce friction by preventing surfaces from coming into contact. ...
... Texture – A rougher surface = greater friction. A smoother surface = less friction. Mass/Weight – The greater the mass or weight of an object, the more friction it will create. Fluids – Fluids reduce friction by preventing surfaces from coming into contact. ...
Hunting oscillation
Hunting oscillation is a self-oscillation, usually unwanted, about an equilibrium. The expression came into use in the 19th century and describes how a system ""hunts"" for equilibrium. The expression is used to describe phenomena in such diverse fields as electronics, aviation, biology, and railway engineering.