Recent content by *Alice*

Thanks a lot - I now have the answer!

Homework Statement The elements of a boiling water nuclear reactor consist of long cylindrical rods of uranium dioxide (U02) of diameter 8mm surrounded by a thin layer of aluminium cladding. In the reactor core the elements are cooled by boiling water at 285°C with a heat transfer...

The questions says: A steel sheet of thickness 5cm is clad on both sides with 2mm thick layers of copper. Determine the stresses and strains induces by a 40K change in Temperature given Data: for steel: E=210kN/mm^2, v=0.3 (Poissons ratio), alpha=11*10^(-6) 1/K for copper: E =180...

Dear all, I am trying to calculate the stresses and strains induced by a 40K change in temperature of a steel sheet of thickness 5mm, that is clad on both sides with 2mm thick layers of copper. For both materials E, alpha (í.e. the thermal expansion coefficient), as well as v (Poissons ratio are...

Thank you. This is what I did: relation 0: b + h(w) = a + h(0) equation 1: 200 = h(0)*850*g This h(0) = 0.024 equation 2: 850g*a = h(w)*g*1000 + b*850*g (a-b)*850g = h(w)*9810 a-b = h(w)*1.18 PUT IN RELATIONSHIP 0: a-b = h(w) - h(0) 0.18h(w) = -h(0) h(w) =...

sorry, I'm confused. What do you mean by height of the oil exactly, i.e. from which point to which? Also shall I call the height of the oil on the left arm h0 or a (you ay two different things in your post). I thought that it was 25 h1, since the volume change should be the same. Since the...

C and D are at the equilibrium levels on the rhs and lhs respectively. Basically I though that the level of the water below the equilibrium level on the rhs must be (50/10)^2=25 times the height difference h1 at the large diameter on the top (see attached file in the above post)

Please see attached the definitions. pressure at C = pressure at D p(C) = p(D) p(D) = 1000g(x-h1) + h(2)*g*1000 + 200Pa p(C) = 850g*25h(1) + (h(2)-25h(1))g1000 + (x + h(1))*1000g now p(C) = p(D) and solve for h(1): 200 - 1000gh(1) = 1000gh(1) - 2500gh(1) + 2150gh(1) h(1)...

Dear all, would anyone be able to help me with the following problem In the manometer shown in the figure, the reservoir diameters are 50 mm and the tube diameters are 10 mm. The upper fluid is an oil of density 850 kg/m3, and the lower fluid (hatched) is water. Initially the levels of oil...

"Electrons are accelerated by a potential of 350kV in an electron microscope. Calculate the de Broglie wavelength of those electrons taling relativistic effects into account" I attempted the following: W = W(kin) = 350keV now W(kin)= (1-gamma)mc^2 so, now one could solve for...

That's exactly what I did and that caused all the trouble: substitute: u= x^2 + a^2 so then you have to multiply by (1/2)...oh yeah...I see! I didn't multiply by two when I did the integration... Oh dear! :cry: Anyways - thank you so much! :cool:

Yes, sorry...missed to write that one in one line. However, I had it back in the integration the line below, so that it didn't affect the answer. It's now edited. Does anyone have any idea about that factor 4?

given: the electric field at a point on the axis a distance x from the plane of a ring is E = \frac {q*x} {4*pi*E0*(x^2+r^2)^{3/2}} where E0 is the permeability coefficient The charged ring is replaced by a circular sheet of charge of radius a a surface charge density sigma. The ring...

cool thanks...so when one now differentiates v wrt T you get a = \frac {g*(1 + alpha*T)} {1 + alpha*t} ok, so T is the time in which the drop falls through the cloud. But what is small t? :rolleyes: It must be a constant as we integrate wrt to T and not t. If t was just big T a=g...which...

hmmmm...ok...so, if I put in t instead of T, I get an equation with two unknowns: v = \frac {gT + 0.5g*alpha*T^2} {(1 + alpha *t) + v(initial)} how can one differentiate that? before I used the following to find a: (U´ * V - U* V´) / V^2 , as well as the derivate of v(initial) wrt t...