Recent content by dandy9

Wow, thanks! That was really thorough and I understand the question now. Thanks for taking your time to help me! I truly appreciate it!

Hmmm... Okay, thanks. But I'm unclear about what to do with that information. What if I do E=mc^2 with m being the combined mass of the proton and the antiproton (so really just 2 times the 1.67 etc.) and get that energy, does this get me anywhere? How do I connect that with gamma rays?

Homework Statement The mass of a proton is 1.67 multiplied by 10-27 kg. (a) Find the energy equivalent of the proton's mass in joules. (b) Convert this value to eV. (c) Find the smallest total g ray energy that could result in a proton-antiproton pair. Homework Equations...

Homework Statement A length of wire extends upward through a piece of cardboard, bends around, forming a loop, and extends back down through the cardboard. All that is visible above the surface of the cardboard is this loop of wire. From your perspective, you are looking downward on the piece...

Yes! Thank you so much. I really appreciate your time and your dedication to help students.

Hmm... Thanks. But am I wrong in thinking that friction = uN and because this is on an inclined slope you can't take the regular m*g, you have to take into account the angle - which is where I got my mg(cos20)... Let me see how it goes.

My equation was: N = normal force u = coefficient of kinetic friction W = uNcosxd W = (4E3)(2.1E3)(9.8)(cos20)*(cos180)(4.7)

Homework Statement A 2.10 multiplied by 103 kg car starts from rest at the top of a 4.7 m long driveway that is sloped at 20° with the horizontal. If an average friction force of 4.0 multiplied by 103 N impedes the motion, find the speed of the car at the bottom of the driveway. Homework...

Homework Statement Three carts of masses 4.0 kg, 10 kg, and 3.0 kg move on a frictionless horizontal track with speeds of 5.0 m/s, 3.0 m/s, and -4.0 m/s, as shown in Figure P6.34. The carts stick together after colliding. Does your answer require that all carts collide and stick together at...

Homework Statement A 64.2kg man fires a .3kg arrow. The arrow is measured to have a velocity of 63m/s by a stationary bystander. All these happen while the archer sliding backwards at -.26m/s. What will be his velocity (in the x-direction) after firing this arrow? Homework Equations...

Ahh. Just kidding. I didn't realize that I got locked out of the question. Perhaps you could tell me if my answer is reasonable just so I know for the future if I did it right in the end?

Does T include f? The way I thought about it was that T is the original tension between the two masses, and then there is an additional force, f, that has to be taken into account. So I'm adding T and f for the "upward force" and subtracting m2g for the "downward force." Hmm. I guess not then...

Still didn't work, but thanks anyway.

Homework Statement Suppose that in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 6.45 kg and m2 = 11.90 kg, what value of f will reduce the acceleration of the system by 59%? Homework...

Thank you! I got them all - thanks for getting me started! This is how I did it: (a) cos and sin are equal at 45degrees (b) the sin of 0 gives a component of 0. (c) if the component and weight are equal then you take the inverse sin of 1 and get 90.