4.3 Acceleration Acceleration describes how quickly speed changes
... Projectile motion has two parts, or components, horizontal and vertical These two components are independent of one another; that is, they have no effect on each other ...
... Projectile motion has two parts, or components, horizontal and vertical These two components are independent of one another; that is, they have no effect on each other ...
ch04
... • Select an object(s) to which the equations of equilibrium are to be applied. • Draw a free-body diagram for each object chosen above. Include only forces acting on the object, not forces the object exerts on its environment. • Choose a set of x, y axes for each object and resolve all forces in the ...
... • Select an object(s) to which the equations of equilibrium are to be applied. • Draw a free-body diagram for each object chosen above. Include only forces acting on the object, not forces the object exerts on its environment. • Choose a set of x, y axes for each object and resolve all forces in the ...
Chapter 5
... 5.4 Length of a Plane Curve • A plane curve is smooth if it is determined by a pair of parametric equations x = f(t) and y = g(t), a <=t<=b, where f’ and g’ exist and are continuous on [a,b], and f’(t) and g’(t) are not simultaneously zero on (a,b). • If the curve is smooth, we can find its length. ...
... 5.4 Length of a Plane Curve • A plane curve is smooth if it is determined by a pair of parametric equations x = f(t) and y = g(t), a <=t<=b, where f’ and g’ exist and are continuous on [a,b], and f’(t) and g’(t) are not simultaneously zero on (a,b). • If the curve is smooth, we can find its length. ...
Chap4-Conceptual Modules
... 10. ConcepTest 4.4c Off to the Races III We step on the brakes of our Ferrari, providing a force F for 4 secs. During this time, the car moves 25 m, but does not stop. If the same force would be applied for 8 secs, how far would the car ...
... 10. ConcepTest 4.4c Off to the Races III We step on the brakes of our Ferrari, providing a force F for 4 secs. During this time, the car moves 25 m, but does not stop. If the same force would be applied for 8 secs, how far would the car ...
Chapter 15– Oscillations
... • (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. • (b) The frequency is the reciprocal of the period: • f = 1/T = 1/(0.500 s) = 2.00 Hz. • (c) The angular frequency ω is ω = 2πf = 2π(2.00 Hz) = 12.6 rad/s. • (d) The angular frequency is related to the spring constant k and t ...
... • (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. • (b) The frequency is the reciprocal of the period: • f = 1/T = 1/(0.500 s) = 2.00 Hz. • (c) The angular frequency ω is ω = 2πf = 2π(2.00 Hz) = 12.6 rad/s. • (d) The angular frequency is related to the spring constant k and t ...
Chapter 7
... A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and th ...
... A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and th ...
chapter12
... When the frequency of the driving force is near the natural frequency (w w0) an increase in amplitude occurs This dramatic increase in the amplitude is called resonance The natural frequency w0 is also called the resonance frequency of the system ...
... When the frequency of the driving force is near the natural frequency (w w0) an increase in amplitude occurs This dramatic increase in the amplitude is called resonance The natural frequency w0 is also called the resonance frequency of the system ...
Honors Work, Energy
... F and d are in opposite directions so = 180 so: W=Fgcos d=-Fgd At the moment the ball reaches the highest point, v=0 so K=0 Wtotal=K and K=1/2mv2 W=K0-K1=0-1/2 mv2=0-1/2(0.06kg)(50m/s)2=75J W=-Fgd d=-W/Fg d=-W/mg d=75J/(0.06kg)(10m/s2)=125m ...
... F and d are in opposite directions so = 180 so: W=Fgcos d=-Fgd At the moment the ball reaches the highest point, v=0 so K=0 Wtotal=K and K=1/2mv2 W=K0-K1=0-1/2 mv2=0-1/2(0.06kg)(50m/s)2=75J W=-Fgd d=-W/Fg d=-W/mg d=75J/(0.06kg)(10m/s2)=125m ...
College Physics
... 4.6 Motion in the Presence of Resistive Forces Consider the effect of the interaction between the object and the medium (a liquid or a gas) through which it moves. The medium exerts a resistive force R on the object moving through it. Examples : The air resistance associated with moving vehicle ...
... 4.6 Motion in the Presence of Resistive Forces Consider the effect of the interaction between the object and the medium (a liquid or a gas) through which it moves. The medium exerts a resistive force R on the object moving through it. Examples : The air resistance associated with moving vehicle ...
chapter 5
... space, the number of lines that intercept the second object falls off with distance as 1/r2 • The result implies that gravity follows an inverse square law because we live in three-dimensional space • It also means we should expect other forces described by the field line picture to have the same in ...
... space, the number of lines that intercept the second object falls off with distance as 1/r2 • The result implies that gravity follows an inverse square law because we live in three-dimensional space • It also means we should expect other forces described by the field line picture to have the same in ...
Physics of Rolling Ball Coasters
... Until now we have considered only straight paths. While these are much simpler, they would make a very boring rollercoaster. Now we need to put together all the theory discussed to this point. Given a parameterized path r(s), define Tˆ , N̂ , and B̂ as the principal unit vectors in the tangential, n ...
... Until now we have considered only straight paths. While these are much simpler, they would make a very boring rollercoaster. Now we need to put together all the theory discussed to this point. Given a parameterized path r(s), define Tˆ , N̂ , and B̂ as the principal unit vectors in the tangential, n ...
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.