Recent content by Robertoalva

yes sin(u+v)=sin(u)cos(v)+cos(u)sin(v) in which this case would be sin(u+(pi/2))= sin(u)cos(pi/2)+cos(u)sin(pi/2)

according to what i have read, sin(u+(pi/2))= cos u but, can I do it with sin(u+(pi/6)) ?

but he said that u= theta - (pi/6) not pi/2

that's how I did it... sin(u)/u= 1 so 1-(1/2)=1/2

?

what i have right now is lim u->0 (sin(u) - (1/2))/u. after that I know that lim u->0 sin(u)/u= 1 then 1-(1/2). but it gives me 1/2 and that's not the correct answer. what am I doing wrong?

so basically the whole thing gives me 1? or the -1/2 stays and subtracts after i do the limit?

Homework Statement Find the limit Homework Equations lim θ→ ∏/6 (sinθ - 1/2)/(θ - ∏/6) The Attempt at a Solution well, if you substitute the values, it gives 0/0, but that means that is undefined and that's not the answer. according to my book the answer is √(3)/2 and I am not supposed to...

well, wouldn't it be, energyrequired to move m1= energyspring-energym2 ?

energy+energy= force?

well, m1gmew is because the gravity acts down on it, mew is against it so everything acts on that fragment of the system, then the m2gh is the same but now with the height actin too, and then the hook's law is just subtracting from the system

1. Two blocks are connected by a light string that passes over a frictionless pulley as in the figure. The block of mass 8.24 kg lies on a horizontal surface and is connected to a spring of force constant 335 N/m. The system is released from rest when the spring is unstretched. If the hanging...

so then it will be, KE+U-W=Total Energy ?

so, i just use KE and U? and how will it be then?

1. A 9 kg block is pushed by an external force against a spring with spring constant 118 N/m until the spring is compressed by 2.1 m from its uncompressed length (x= 0). The block rests on a horizontal plane that has a coefficient of kinetic friction of 0.67 but is NOT attached to the spring...