Recent content by ledhead86

so does conservation of momentum not apply her? What do I need then to find the magnitude of the recoil velocity?

No, there are no external forces. Initial momentum= mv= 19233429*6.65*10^-27= 1.27902306*10^-19

Anyone Know?

The nucleus of 214Po decays radioactively by emitting an alpha particle (mass 6.65*10^-27) with kinetic energy 1.23*10^-12 J, as measured in the laboratory reference frame. Assuming that the Po was initially at rest in this frame, find the magnitude of the recoil velocity of the nucleus that...

Thank you whozom for completely wasting the past two hours of my life.

NO I CANT! .5v^2=gh .5v^2=931 v^2=1862 v=43.15 = INCORRECT ANSWER!

Sorry, I just don't understand this at all, but I already said from the beginning that .5mv^2=-mgh, but I still don't know how I am suppose to solve for v without m.

so pe=g*95 ?

Not Really. The potential energy at point a= mgy = m(9.8)(95) does it not? I still don't know the mass.

How would I set that up if I don't know the mass or weight of the coaster. The only equation I know is (1/2)*m*v^2 + m*g*y_1 = (1/2)*mv^2 + mg*y_2

Any help would be appreciated

Riding a Loop-the-loop. A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle. http://community.webshots.com/user/mmaddoxwku" If the car starts at...

Never mind, I figured it out.

OK. So I know -delta_U=F* D * sin(theta) So. 9.68=F*D*sin(37). What is F. Is that 400N, the spring constant?

When compressed the block has potential energy and no kinetic energy, and when released the potential energy will be converted into kinetic energy because now the block will be moving. Since the sufrace is frictionless the block will move at a constant velocity having a KE equal to the initially...