=Homework Problem: Longest Vertical Soda Straw

AI Thread Summary
The homework problem focuses on determining the maximum height of a vertical soda straw that can be used for drinking. Key concepts include understanding atmospheric pressure, cohesive and adhesive forces, and the mechanics of sucking liquid through a straw. It's important to realize that when sipping, the liquid is pushed up by atmospheric pressure rather than being pulled. Clarification on the question's intent is needed for better guidance. Overall, grasping these principles will help in solving the problem effectively.
Gott_ist_tot
Messages
52
Reaction score
0
I received a homework problem that I need guidance on where to start.

What is the longest vertical soda straw you could possibly drink from?

I do not know where to start on this problem. A small push in the right direction would be greatly appreciated.
 
Physics news on Phys.org
Can you please specify your doubt a bit more. I didn't understand what the question means?
 
Probably something to do with cohesive and adhesive forces? or over coming gravity by the force exerted by the sipping the straw
 
Think about what's actually happening when you suck.

You're not so much pulling liquid up the tube as much as allowing atmospheric pressure to push it up.
 
Oh, and please don't repeat your questions. I've just seen that you had a good answer given just a few hours ago, the first time you asked.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top