Recent content by destro47

The light doesn't "know" which path/time is the shortest...The fact is that light has wave characteristics as well as ray properties. The light waves propagate along ALL paths...the waves that are traveling along paths that are not minimized will reach the point exactly out of phase and undergo...

the left side of the equation is Joe and the right is Tom

1/2 mv^2 = 1/2 mv^2 1/2 m (.228v)^2 =1/2 mv^2 1/2 m .052v^2 = 1/2 mv^2 .052 (89.7) = m

Having trouble with the following two part problem: A) Joe, mass 89.7 kg, is racing against Tom. When Joe and Tom have the same kinetic energy, Tom is running faster. When Joe increases his speed by 22.8%, they are running at the same speed. Find Tom's mass. Tried equating their kinetic...

My intuition was correct. 1.68 N is the measure of the tangential force, calculated using the acceleration of the car over that particular arc length of the circle. 0.55 N is the measure of the centrifical force keeping the toy car on the circle.

so basically the answer is -706 N if i understand you correctly: Pulling Force + Friction Force = m (0) 706 N + f = m (0) f = 0 -706 N

Sorry but my class has not covered W yet. I believe its W=fd, right? My professor is a real ball buster who genuinely enjoys making his students squirm over the homework assignments. Any other suggestions?

The biggest problem that I had with this problem is that the log moves at constant v, so a=0. How can a force move on object with 0 acceleration? Is this log considered to be in equillibrium even if its moving?

One end of a horizontal spring (k = 333 N/m) is attached to a 3.12 kg box, and the other end to a fixed, vertical wall. (A picture of this situation can be found in Fig 4-10 on page 102 of your textbook.) a) Find the magnitude of the force needed to move the mass so the spring is stretched...

A red box and a blue box sit on a horizontal, frictionless surface. When horizontal force F is applied to the red box, it accelerates 4.88 m/s2. a) If F is applied to the blue box, it accelerates at 1.32 m/s2. What is the ratio of the masses of the boxes (mass of the red box/mass of the blue...

Then how does it apply to this problem: A 7.83 kg toy car is going around a circular track of radius 52.5 m at a constant speed of 17.3 m/s. Find: - the time it takes for the car to go around the track once - the magnitude of the inward force needed to keep it moving in a circle...

A rope is tied to a 97.8 kg log. When you pull on the rope with a horizontal force of magnitude 706 N, the log moves at constant speed. a) Find the magnitude of the resistive (frictional) force acting on the log. You may assume the frictional forces acting on the log is constant. N b)...

In Uniform circular motion, the centripetal acceleration is the inward force that keeps a particle on a circular track. My question is what exactly is the tangential force? Is it a fictious force? My first inclination is that is equal in magnitude to the centripetal force but acts...

never mind...makes sense now. I feel like a total retard. I used Pythagorean theorum to solve part A. I just realized that the pythagorean theorum is becomes a special case of the law of cosines when angle C = 90*

If I think that I understand you correctly ur saying to use the law of cosines to find the magnitude of the resultant force graphically. That makes sense, but what I don't understand is how that would work. Only the acceleration is given, the magnitude of the force is not given in this problem...