Recent content by Waffle24

Ah that explains it! Thank you! :biggrin:

I'm not quite sure what you're referring with "if I can satisfy all the conditions". Anyway, the steps to get to the deflection equation is also a little bit confusing to me, because there is a '-' sign before 1/EI. I've tried getting the C2 by substituting the M(2) = -5 in the M equation. Is...

Sorry for not replying to the questions. Meanwhile I've managed to find the answers to the question through the help from my lecturer. My apologies for bumping the topic, but I'll share my answers in here in case anyone is interested. The advantages: ● No space required behind the dikes. ●...

[Moderator moved this thread from the homework section.] I'm a civil engineering student, and it's my first year at the University of Applied Sciences, and I'm doing an exercise but can't figure out one of its parts. The question: what are the advantages and disadvantages of reinforcement...

Maybe there's a formula or rule that I'm not aware of...:rolleyes:

Right, thank you so much for taking the time to help me. :smile:

On edit : Ah I see it now, so the difference of leg 2 is 15cm Hg. Total = 3cmHg(leg 1) + 15cmHg(leg 2) = 18 cmHg has drained. Now you can get the Volume : V = 2 × 18 V = 36 cm^3 :smile:

Ah, I think I've misunderstood the question "What is the difference in height of the mercury columns in legs 1 and 2?" So I actually need to tell the difference of each of column right? In leg 1 : 19cmHg - 16cmHg = 3 cmHg In leg 2 : Eh, would you like to help me on this one? Since I don't...

Well it has moved, because the air pressure was higher than the trapped air. 12cmHg So it will be : 19cmHg - 12 cmHg = 7 cmHg

Sorry for the late reply. V = 2 x 16 = 32 cm V = 2 x 19 = 38 cm P1 x V1 = P2 x P2 76 x 32 = P2 × 38 2432 = 38P2 P2 = 64 cmHg Trapped air pressure = 64cmHg Air pressure Difference between leg 1 and 2 : 76 - 64 = 12 cmHg What is the difference in height of the mercury columns in legs 1 and 2...

I'm stuck at here, like is there an equation to calculate the new pressure(trapped air)? On edit : I think I've figured it out already, since both length and the area of both column are known, we can calculate the Volume and then apply Boyle’s Law.

Well the pressure of the air in leg 1 is equal to the air pressure if both K1 and K2 are closed, so if you open up K2 the air pressure in leg 1 will be less than the air pressure in leg 2, and so will the liquid drop faster in leg 2, because the air pressure in leg 2 is bigger than the air...

Homework Statement The open legs 1 and 2 of an U-shaped tube have a diameter of 2 cm^2. In leg 2, mercury is poured. When the distance is 16cm from the mercury level up to the valve K1 , then the valve gets closed. The barometer reading is 76cm Hg. a) How big is the pressure of the sealed air...

Ah that explains it, thank you sir for your help. :smile: