Recent content by shotgunbob

Homework Statement A lunch tray is being held in one hand, as the Drawing illustrates. The mass of the tray is .2 kg and its center of gravity is located at its geometric center (the tray is .4m long). On the tray is a 1 kg plate of food and a .25 kg cup of coffee. Obtain the force T exerted...

whoops! Sorry! it should be: If 5 employees decide not to take any tickets, all the other employees will receive 7 tickets, and none will be left over.

Homework Statement A small company has some free movie tickets to distribute to its employees. If each employee gets 5 tickets, there will be 15 left over. If 5 employees will receive 7 tickets, and none will be left over. How many tickets does the company have? Homework Equations...

Homework Statement You are at the controls of a particle accelerator, sending a beam of (1.5x10^7) m/s (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such...

Thank you!

I suppose its the top of the spring

Ok I got .54 m when i put Pe(grave)=Pe(spring). Thats an answer choice! Is this correct?

I messed up, used PE=1/2kmx^2 for some reason. I got 1.6 J. And in Grav. PE, the H should be the same as the H I am looking for so PE(grav)=PE(Spring). Or maybe I am missing something.

Homework Statement A .3 kg mass rest on top of a spring that has been compressed by .04 m. Neglect any frictional effects, and consider the spring to be massless. Then, if the spring has a constant k equal to 2000 n/m, to what height will the mass rise when the system is released. Homework...

This is essentially the same type of question i asked. You have to isolate tension one from T2 and substitute. I am lost from then on lol

hmmm t1= t2 (cos 45)/cos 30) So now I have T2((cos45)/(cos30))-T2Cos45 For which I get .8164965809 hmmmmm where did i go wrong

Basic Tension question- 5 kg block hanging from a string with one side at 30 degrees and the otherside at 45 degrees. Fnet,x=t1x-t2x=0 Gravity assumed at 10 m/s2 (in the book) I got to: T1(cos 30)-T2(Cos45)=0 for the total forces in the x direction and T1(sin 30) + T2(sin 45)-(50 N)= 0 for the...