Recent content by Newton=boss

Of course, if you don't believe Newton's second law either, then that argument is no good.

This is not true. My explanatory abilities are limited, but think about it this way. If for example you jump in the air, you are applying a force on the earth, and the Earth is applying a force on you. Lets assign this force an arbitrary value..30 N. F=ma. Let's say you have a mass 60kg...

check that. I was able to solve part a)..thankyou! I do not even know how to start on the next 3 parts..if you could point me where to go I would be indebted. wouldnt b simply be to divide the rotational Kinetic energy by the total kinetic energy? apparently not..

the rotational kinetic energy can be found from I=1/2*40*4*.4^2? and 60*1000/3600=v omega=v/0.4?

Homework Statement The 1900 kg mass of a car includes four tires, each of mass (including wheels) 40 kg and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder. (a) Determine the total kinetic energy of the car when traveling 60 km/h. (b) Determine the...

ah! thankyou

I don't know where I am wrong..

1.6=1/2(4.7)(0.078)^2*alpha alpha=111.908 rad/sec^2..this is a negative quantity Omega initial=5000 rpm *pi/180=87.266 rad/sec Omega final=0= 87.266^2 +2(-111.908)(Theta) Theta= 34.02 rad*180/pi = 1949.51 revolutions

1.A centrifuge rotor rotating at 5000 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.60 m·N. If the mass of the rotor is 4.70 kg and it can be approximated as a solid cylinder of radius 0.0780 m, through how many revolutions will the rotor turn before...

Its deceptively simple!

oh and F frictional=mu*Normal Force

I have the same question, and the relevant equations for the section are Energy equations and possibly momentum equations. K=.5mv^2=Fd m1v1=m2v2 however, like the OP, as the stone is at rest and has zero net acceleration/ force I do not know which equation to even use. If anyone caould point...