yes sorry, you're right... it was meant to read 49.7cm, which i have now changed! sorry! i have still been trying to do it, found a number of various formulas and ways to go about it on the net, but none are helping me to find the mass of the stick, rather the masses of the objects joined to the...
ahhhhhh yes you're quite right! the stick balances at the 49.7cm mark, for some silly reason i decided to write the 24.7cm mark... mmm! i think i might just have to wait to find out how to do it in my tutorial, i have no idea what formulas to use or anything, thanks for your help, but i think i...
well the center of mass of the meter stick is a little off half way along the stick, but a meter stick can not be heavier on one half then on the other, and there are no masses attached to either end (originally) so i don't know how i am meant to find the mass of the stick with such limited...
a metre stick is found to balance at the 49.7cm mark when placed on a fulcrum. when a 50g mass is attached at the 10cm mark, the fulcrum must be moved to the 39.2cm mark for balance. what is the mass of the stick?
==> i have NO idea what to do with this quesiton... have tried various ways to go...
3) a metre-stick is found to balance at the 49.7cm mark when placed on a fulcrum, when a 50g mass is attached at the 10cm mark the fulcrum must be moved to the 39.2 cm mark for balance. what is the mass of the meter stick?
- just plain stuck on this, tried about 3 different ways of attempting...
thanks
ok, i think i have got this one... i was just using the wrong value for force, which was throwing it all out.. once using -2m/s^2 i put that into the v^2=v(initial)^2 + 2a (x-x(initial) or something along those lines... it then came out as 24.5m/s, which fits with the rest of the...
2) the driver of a 600kg sports car, heading directly for a railroad crossing 250m away, applies the brakes i a panic stop. the car is moving at 40m/s, and the brakes can supply a friction force of 1200N. a) how fast is the car moving when it reaches the crossing?
- i tried finding the change...
ok... i think i have got it... i followed what you and skippy said to do, used Distance= v(initial)t-.5gt^2 with v(initial) as 9.8 and time as 1.00s... i came up with 4.9m. this was also true of a different equation that i found online, similar idea and same answer... YAY! thanks for that...
maximum height??
1) A ball is thrown vertically upward and is caught by the thrower after 2.00s. find a) initial velocity b) the maximum height it reaches?
a) initial velocity obviously = o, but i have no idea how to find max. height?
2) the driver of a 600kg sports car, heading directly...