Recent content by mh1985

T=2pi(sqrt(l/g) ?

Homework Statement Pendulum of unknown length, l, has period of 9.32 seconds. Length is increased by 1 metre, and time period increases to 9.734 seconds. Calculate original length of pendulum and determine whether the experiment was conducted on earth. Hint: g is not assumed to be 9.81 ms^-2...

can anyone help?

Homework Statement If in an open water channel, the approach speed of an incompressible ideal fluid, of constant depth h1, is not too large, a small bump in height H (H<< h1) in the base of the channel will cause a dip Δh (Δh << H) in the water level directly above it, such that h1 = Δh + H...

So I have: P1* (P2T1/P1T2)n= V2 ?

Hi thanks for the reply If I make V1 the subject, I get P2V2T1 /P1T2, right so far? :S But I don't see how to substitute it into the first equation? P1 * (P2V2T1 /P1T2) = P2V2^2

Thanks for the reply, So I multiply both sides of P2/P1=15 by P1 to get P2, 101325 * 15 = P2 = 1519875 Pa Not sure how to form the equation involving only P & T... Something like T2/T1 = (P2/P1)^((n-1)/n) EDIT: so (P2/P1)^((n-1)/n) = 15^(1.1 - 1)/1.1 = 1.2791 1.2791 = T2/T1 T1*1.2791 =...

Homework Statement Using a value of n of 1.1, calculate the final pressure and temperature when the cylinder is compressed rapidly with a compression ratio of 15. Starting pressure = 101.325 kPa Starting temp = 298 K Homework Equations (V2/V1) = 15 The Attempt at a Solution...

Why can there sometimes be a difference observed between them?

I have just seen it now, I was expanding the brackets out before, but if I don't do that, on the 2nd equation where we have 5(x-2.5) for x=2.5, this is 5(2.5-2.5) = 0...must've made an error when expanding them out. thanks!

Thanks, well at least I have the right answer. I think I'm doing something wrong with the bending moment equations in that case because I get something different for each one: for 1 < x < 2.5 M = 12.98x - 0.0951x^2 - 10(x-1) x = 2.5, M = 16.86 kN*m for 2.5 < x < 4 M = 12.98x -...

many thanks for pointing those out! With regard to the moment being different depending on which equation I used, what about when I use the one for 2.5< x <4, shouldn't this give the same value as the equation for 1 < x<2.5 provided I used x = 2.5 ? I've redone it and come up with this...

but aren't there two depending on which side of 2.5 x is? ie. 31.855625 kNm and 18.855625 kNm EDIT: I have done tau_max = Mc / I for both: (31.85 x 10^3 N x 0.125 m)/ 2.1484 x 10^-4 m = 18.53 MPa (18.856 x 10^3 N x 0.125 m) / 2.1484 x 10^-4m = 10.97 Mpa So the beam cannot support these...

Thanks. how do I include the weight of the beam in this though? Which value for M do I use in the bending stress equation?

Not sure if this is right but: Bending moment at x = 2.5, (12.98 * 2.5) - (10 * 1.5) + (10 * 1.5) - (12.98 * 2.5) = 0 ?? Again, don't know if I need to do this but I have sectioned the beam and done the bending moment diagrams, which has given me the equation: 12.98x - 0.0951x^2, for x= 2.5...