Friedberg's Linear Algebra states in one of the exercises that an inner product space must be over the field of real or complex numbers. After looking at the definition for while, I am still having trouble seeing why this must be so. The definition of a inner product space is given as follows...
This confuses me. I usually think of heat as energy being transferred from one thing to another. Can we really speak of a transfer of energy in all chemical reactions? Suppose we have a reaction of two solid chemicals that produces gas at constant temperature. The enthalpy for such a reaction...
Lately I've been struggling with the idea of enthalpy and what it means conceptually, especially in its use in Gibb's free energy. There is nothing in the definition of change in enthalpy that would connect logically to spontaneity or free energy.
After thinking about it for a couple days...
No. First, it would be v_{f}=\sqrt{2gh} but you can't use that equation anyway because in addition to translational kinetic energy, there is rotational kinetic energy.
Start off by looking up or deriving the moment of inertia of the cylinder for the rotation. Use your conservation of energy...
Friction for a person walking is static friction because there is no relative motion between the surfaces. Friction for a breaking car depends on what the car is doing. Most of the time when you are acclerating (or decelerating) it is static friction because there is no relative motion between...
The the maximum height, its vertical velocity is zero. It has only horizontal velocity, and horizontal velocity depends on the velocity at which it was fired. It isn't possible to solve this problem with the information you are given.
Oops. For some reason I thought you did your free body diagram the same way I did. I have my free body diagram set up a bit different so the origin is at the bottom. It doesn't matter in any case. Where you set it won't affect your final answer.
The force of friction and the weight will tend...
Yes. Now get your torque equation set up so you can find the position of the person. It is most convenient to use the right end of the ladder as your origin.
Yes. It should be clear from your free body diagram that the y normal force minus the combined weight of the ladder and person gives the net force acting in the y-direction. Acceleration is zero so the y-normal force equals the combined weight of ladder and person.