Recent content by Arty7

so.. mgh+1/2mv^2 0.5 +32 =32.5J

sorry i don't understand that very well.

The question is in the attachment. a) for the height i got 1.25m which you get by using displacement and acceleration. (t=20/40=0.5sec) b) a=v/t, 10*0.5=5m/s c) At initial point: mgh= 0.040*10*1.25=0.5J 1/2mv^2=1/2*0.040*40^2=32J Hits Ground: mgh=0.040*10*0=0J 1/2mv^2=1/2*0.040*(20 or 40)^2...

Sorry man, its up now

IHi guys with part b of this question I am having trouble. I have cut it it at the specified members and used the top half. Now i take the moment at A to get the force in BC. (CW+)M@A=(6*3)+(6cos(30)*3)=0, But when you solve this it doesn't give you the answer its supposed to be, I am probably...

Homework Statement Please check the attachment to see all the variables Homework Equations Xl=2*pi*f*l, Xc=1/2*pi*f*c, variation of ohms law The Attempt at a Solution Ok for 5) i found Xl & Xc, then i combined the resistances in parrallel and used ohms law to find I=v/r. My...

hi guys i got i a practice exam of teacher yesterday, its been really easy but i don't understand this one question. Discuss that how kirchhoff’s voltage law is applied in the electrical transmission system?. I know this is a easy question but i don't understand how kirchhoff’s voltage law...