Recent content by ixbethxi

i concluded that it waas the cable that was defective

honestly i tried understanding what you said, maybe I am just slow. i understand your last comment about how making them equal is completely wrong, yes that makes total sense. but if its possible can u explain to me the part about the q and the cos...you completely lost me. sorry

my work is what i showed above, I'm trying to see if that's right, but the picture is still pending aproval, I am sure u'll understand as soon as its approved.

oo its basically the picture.. "a small ball rolls arond a horizontal circle at height y inside a frictionless hemispherical bown of raidus R as shown in my picutre find an expresion for the balls angular velo in terms of r,y,and g

find an expression for the balls angular velocity in terms of R,y,and g mr*omega^2=mg so i had omega=sqrt(g/R) and I am saying that the angular velocity does not depend on its height. am i right?

i mean let's say you are holding the mass on the incline still and then you let go, does the mass that's heavier determine if the mass on the incline will make it slide up or down?

oo i forgot to say its frictionless ;)

i just was wondering, when you have an incline and a box mass m sits on it attached to a massless string and a pulley and a weight hanging freely. what determines whether the block will slide up or down the ramp? is it the horizontal weight force for the box and the force of gravity for the...

You've been called to investigate an accident in which a cable broke while lifting a 4500kg container. The steel cable is 2.0cm in diameter and has a safety rating of 50,000N. The crane is designed not to exceed speeds of 3.0m/s or accelerations of 1.0m/s^2, and your tests find the crane is...

ok this is what i got so far 0.5*m*v^2= 0.5*.290kg*200cm/s^2= 5800 5800= (0.5)(.110kg)(v^2) i solved for v and i got 324.7cm/s^2 now i know this is the velocity at the bottom of the ramp but i don't know hwo to get the height from here.

o wow yea that makes a lot more sense than what i had, i used the wrong eq. lol thank youuu!

o whoops.. i meant cm/s

A 110 g marble rolls down a 39.0 deg. incline. At the bottom, just after it exits onto a horizontal table, it collides with a 290 g steel ball at rest. How high above the table should the marble be released to give the steel ball a speed of 200 m/s? well first i tried to use the formula...