Recent content by bionut

YEs, I thought of that... but don't know the mass of the ball?

Hi all, I know this may see basic but how would I go about calcualting Impact velovity? The only information I have is: Height dropped, height bounced and from here I can calculate the CofR... But how would I go about finding impact velocity

Calculate the pull on the high bar of a gymnast performing a giant swing. His angular velocity is 2 rads, his mass 80 kg and effector radius of rotation (distance from bar to CofG) is 1 meter Hi... I am a bit confused... "pull on the bar" would this be the work done on the bar to the person...

So if he's moving at a constant velocity --- he's not accerating therefore threre is no force affecting his velocity?

An athlete (mass of 50 kg) is moving with a constant velocity of 5ms. Determine the resultant force acting on him. If it's a constant velocity he is not acclerating? Therefore a = 0 m/s/s; F=ma F=50 X 0 = 0N? But there has to be a force dosent there? Would the force be equal to the mass...

yes... get it now... fwwwww lol... thanks for all you help!

(Q) What torque is applied to a steering wheel of an automobile when the driver applies a force of 50 Newtons tangent to the circumference of the wheel? Diameter of the steering wheel is 40 cm. F = 50 N D (torque arm) = 0.40m T= Fd = 20 N... but its wrong... any ideas...

Yes but it says what is her maximum velocity: So v=sqrt 2gh @ 1m = 4.43 m/s but v=sqrt 2gh @ 2m = 6.26 m/s ... (so this does not make sense to me...)

So then this indicates that the max velcoity is at the higest point? I though max velcoity is at the lowest point??... but her velocity @ 1m using v= sqrt 2gh also = 4.43 m/s... so is the velcoity then constant?

okay well.. the answer is 4.43 m/s... so v=sqrt (2 x 9.81 x 2) = 4.43m/s... aff yeas on reading the question it say's her velocity at highest point... Thanks for all your help... I undersatnd how to getthe answer ..but confused how it can be solved without a mass...

rayquesto ... but this would make here velocity at the highest point faster than the lowest point... My undersatdning is velocity is at it/s max when d=1m...

Okay, if PE = 9.81 J @ 2m above the ground... and energy is conserved, @1m above the ground all energy will become KE?; KE=9.81J? if I plug that into KE=1/2mv(sqr) I still can't comput an answer... I don't have mass (sorry... this is fustraing the ! out of me lol)... I understand...

Hi all thanks for all you help... PeterO... the equations 1/2mv(sqr)= 0 = KE @ 2m? and mgh=0 = PE @ 1m? so.. PE + KE = C would be: mgh + 1/2mv(sqr) = c? im really confused... I don't think the question is intended to be this complicated... I think its trying to gte me to think about PE...

All I can think of is the realtionship between energy conservation... So at 2m (max height ) Her PE = max and at 1m (min) her KE=max where her velocity is max and her PE = min/0j) If I use PE=mgh at 1 m; 0=m x 9.81 x 1 = 9.81 kg... Then KE at 2m = 1/2mv(sqr) = 0=4.91v(sqr) = 2.2m/s...

The only issue i have using an energy formula is I don't have the mass? othwise I would use: PE + KE = C wt x h x 1/2 mv(sqr) = c then I would sub in C into: wt x h x 1/2 mv(sqr) = c, to find V...