Recent content by pawlist

Its the same question no mass or forces are given just the diameter ,angular velocity and the velocity we have just found. And g=9.81ms-2 presumably

I used the equation r=v^2/g 17.55^2/9.81=31.4m the answer seems bit big for an arc of a bridge what do you think?

And in another section it says that the motorcycle crosses a bridge just without losing contant with the ground Calculate the radius of curvature of the bridge

I think its because i had my fladh on

Ohhhhhh 3600*17.55=631800/1000=63.18km/h-1 am i correct?

So now i moved on to the second question and i did v=wr v=65*0.27 V=17.55ms-1 But it needs to be km/hr and the answer must be around 63 However m to km *1000 so 1000*17.55=17550 To get seconds to hrs you do /3600 So 17550/3600=4.875 What ami doing wrong?

So basically w-constant v-variable?

But doesn't the outer part of the wheel cover more diatance than the inner part.It is true that they both take T to complete 1 revolution but the outer part is covering more diatance

So what you are saying is that w is 65 rads-1 at all points?because thr question is asking me to find w at half the radius

I don't know if its correct since i don't have an answer sheet

Homework Statement angular velocity about the axle = 65 rads^-1 diameter = 0.54m radius= 0.27m 1/2radius= 0.135m[/B]Homework Equations v=rw w=θ/t a=v^2/r=w^2r (Might be the probability that i have to use another equation)[/B]The Attempt at a Solution Firstly i cut the diameter in half which...

My final answer is 5.57^-15 can you confirm please?

These are all of the available equations but i don't know which one to use.I uploaded an image

Homework Statement I need to find the magbitude of charges Lenght of string - 50 cm mass of balls - 0.2 grams theeta - 37 degrees Homework Equations K-8.99x10^9 [/B]The Attempt at a Solution I found that half the distance between the balls is 15.9 cm but i don't know how to continue[/B]