Recent content by LBP

Thank you so much! I had a quiz on this yesterday and I am pretty sure I nailed it! So thank you again!

Homework Statement I am trying to figure out how the formula for the force between two anti-parallel wires differs from the formula for parallel wires. Homework Equations B=μ0I/2∏R Force between two parallel wires = μ0LIaIb/2∏d The Attempt at a Solution I understand that d is...

Homework Statement A 0.060-kg handball is thrown straight toward a wall with a speed of 11 m/s. It rebounds straight backward at a speed of 8.6 m/s. (c) The rebounding ball is caught by a player who brings it to rest. During the process, her hand moves back 0.40 m. What is the impulse...

I was afraid of that haha. I appreciate your help anyhow. I have a better understanding of what's going on then when I started, so thank you!

Oh I am getting careless with units! -5.62 cm not meters sorry! The question is from a study guide for which we were given the answers.

humm Okay so I reworked the problem. Knowing the left mass is .1M in from the left and the right mass is .08 M in from the right and the middle is .75M. Then X1= -.65 X2=0 and X3=.67 So I did ((-.65*60)+(0*5)+(.67*40))/(40*60+5)= -.116 but I know the answer is -5.62.

I was thinking subtracting from 1.5 would give me the position on the bar and from that figure out that position's distance from the center, no? Where does the weight of the bar go? Do I add the weight to either position? or do I use its center of mass an weight as a M3X3?

Earlier I meant to say there would be no Static friction, not any friction at all. Is that correct? The two forces are Tension or pulling force and kinetic friction. They must add to 0 if there are no other forces, meaning they are equal to each other?

Homework Statement A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left, and two weights of the same mass are hung on the right. The width of each weight is 4 cm and the outside edge of each group of weights is placed 4 cm from the ends. Where is the center of...

So to make sure I understand this: no net force doesn't mean there is no movement, but rather no acceleration. And since there is no acceleration there is no "force hump" to get over so there is no frictional force. Then since forces have equal and opposite reaction forces the velocity is equal...

It would seem that there are no forces acting on the block if there is a constant velocity.

Homework Statement A block of wood is pulled by a horizontal string across a rough surface at a constant velocity with a force of 20 N. The coefficient of kinetic friction between the surfaces is 0.3. The force of friction is A) impossible to determine without knowing the mass of the block...