Recent content by Bolter

If I take it as a differentiator instead since its a first order differential eqn, then I get this as my new expression for V_out Does this seem about right or not? Sorry if the image is a bit hazy

So I'm I wrong to assume my equation in the first line to be incorrect? This is how I reached to that equation

ok thanks, I was thinking there would be a set value at the end as the end result so it threw me off to begin with.

Hi everyone, I've been having trouble on how to answer this question This is my attempt at it below, but I am not sure what values of t to sub in for evaluating the integral. Any help would be appreciated! Thanks

Thanks I think I get it now. It made sense to give Q (heat energy) a negative sign since energy was dissipated. So the negative value indicates a loss in heat energy. Then working from there I got intial temp to be 428.81K So it should've been (17+273.15) * (–99,000/1.4*510) = 428.81K

Hi everyone This is a quick Q but I don't understand why I got it wrong This is what I have done Could anyone let me know where I went wrong? Thanks for any help!

Hi, Can anyone let me know if I had done this Q correctly? Thanks for any help!

By that do you mean the perpendicular distance from C to BF which is equivalent to length AF as they are parallel to each other? EDIT: Ok I got distance from C to BF to be 1.414m and assuming that's equal to AF then AF = 1.414m hence moment is 1.414m * 50N = 70.7Nm

I tried including line AC in my diagram but I don't see how it'd help me find any angles or lengths. I thought I could somehow use Pythagoras theorem to the right angle triangle ABF (in my diagram) where I know AB is sqrt(10) but I still have 2 unknown sides so that won't work

I redrew the diagram and and labelled the new perpendicular distance I have to find, however I don't think I have enough info in the question to be able to work out this distance unless I'm missing something?

I have tried this question but can't get my head around to doing part b) This is the sketch I have drawn and I'm guessing 'x' is the distance that I have to fine that's perpendicular to force Fb, though I'm not sure if I have set it up correctly? Any help would be appreciated! Thanks

I actually managed to find the right answer at the end by drawing triangles and labelling the forces that are perpendicular to the line AB. Then with some basic trig I got the force needed

Hi I'm very stuck on what to do for these 2 questions I got wrong Can someone please help me on what triangle I need to sketch out in order to find the 2 components of forces for F1 and F2. I'm assuming you have to make use of the sine or cosine rule here I'd be grateful for any help given...

So do I assume that the inlet has 1 m^3/s flowing in and both outlets at the end have 0.5 m^3/s flowing out

Made an attempt at this Q but I'm unsure on how to do part c) or if I had even done part a) and b) right What difference does it make to the volume flow rate equation when the valve goes from open to closed? Any help would be appreciated! Thanks