Recent content by Xenix

[b]1. I need to know the stress σ in the "y" direction. [b]2. I will use σ=P/A equation. [b]3. First (-2*10^6)N*0.4m=-0.8*10^6N Then σ=P/A=(-0.8*10^6N)/(0.4m*0.02m)=-100*10^6Pa. Is this corrects?

Thanks again for the answer. It turns out there is a scheduling conflict at the uni and we are not supposed to cover that topic until next week, however the assignment is due Monday. I managed to solve this though. Thanks for your help! :)

Ok, thank you for your answer, so then 0.1*(4*0.8)?

I am given this probability table: x 0 1 2 3 4 P(X=x) 0.8 0.1 0.05 0.03 0.02 X is the amount of faulty products produced in a day. I am being asked to find the probability of a exactly one product being foulty in a 5 day periode. I am a bit confused. I know from the table that the...

Hi! Thanks! (1375*cos(x))2 + (1375*sin(x))2 = 1 But how does it help me?

[b]1. So I have 2 equations: (1) A*cos(30)-1375*cos(x)=0 (2) 1375*sin(x)+A*sin(30)-1500=0 I redid equation (1) into A=1588*cos(x) And substituted it into equation (2). The result: 1375*sin(x) + 794*cos(x) = 1500 Can someone please tell me how to solve for x please?