View Full Version : PHysics HELPP

01-26-2011, 08:21 PM
A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 17.8 deep.

How far upward from its initial position does the mercury in the right arm rise?

Help me preash

Uncle Flappy
01-26-2011, 08:27 PM

01-26-2011, 08:42 PM

01-26-2011, 08:43 PM
17.8 what? cm?

(1/2)*17.8 cm/13.6 = x

Find x.

13.6 is the specific gravity of mercury.

The added weight of water on one side will then be balanced by the higher column of Hg on the other side. The 1/2 factor is there because it rises on one side and falls on the other.

I'm in Electrical Engg. So.. I could be wrong :)

01-26-2011, 09:26 PM

01-26-2011, 09:27 PM
I assumed so.

01-26-2011, 09:49 PM
it was 6.54mm.

01-29-2011, 02:06 AM
missed this thread earlier but disoblige is right. The way I think of it is you have to figure out what weight of mercury will balance the weight of the 17.8cm column of water. if you write out an equation with the weight of the water on one side and the weight of the mercury on the other side you will see that everything cancels but the height of the mercury


pi *r^2 *(0.178)* 1000kg/m^3* 9.81m/s^2 = delta_h*pi*r^2*13534kg/m^3*9.81m/s^2

(note: i've changed the height to meters to stay consistent with units and i've used the densities of the two fluids to get the masses and subsequently forces by multiplying by g)
the pi's, r's (radius of tube) and accell. due to grav. will cancel and you'll be left with just h as your variable.

then the height must be divided by two because you've had an increase of mercury mass on one side and an equal decrease of mercury mass on the other side defined by a change in height, so h/2 will give you the height change.

This is basically the same thing as disoblige has stated but it doesn't hurt to hear things from two perspectives

note: when I say weight I mean m*g

02-12-2011, 12:14 PM
4.201023 nitrogen molecules collide with a 14.0 wall each second. Assume that the molecules all travel with a speed of 360 and strike the wall head on.

What is pressure on wall in Pa

02-12-2011, 02:10 PM
You couldn't even put in the effort to put in units and you want us to give you an answer?

MasterPhysics is easy man, just plug in formulas and look in the book. But if you want us to help, at least put down some units.

02-12-2011, 04:27 PM
dude just copies and pastes the question without realizing the units are in picture form. nub

02-12-2011, 04:28 PM
Maybe buddy should do his own homework...

02-12-2011, 05:43 PM
Post your masteringphysics username and pass, we'll take care of it for you :thumbsup:


02-13-2011, 04:20 PM
LOL @ no units

02-13-2011, 04:28 PM
Just copy/pasted so units didn't show up . Way too lazy to figure this stuff out, but thanks for all the hate haha.


02-13-2011, 04:32 PM
You'll get far in life for sure hoping other people will do your work for you............

02-13-2011, 04:55 PM
^ its just 2 intro physics questions. Don't think posting them up on a thread hoping for someone to give some hints will cause a determent to my future in academics.

02-13-2011, 05:06 PM
If they are intro questions shouldn't you hope that you can figure them out on your own then?? :dunno:

02-13-2011, 05:43 PM
Originally posted by Unknown303
If they are intro questions shouldn't you hope that you can figure them out on your own then?? :dunno:
give the kid a break!!

02-13-2011, 06:09 PM
Originally posted by tbomb
Just copy/pasted so units didn't show up . Way too lazy to figure this stuff out, but thanks for all the hate haha.

lol.. Best of luck then. Don't be surprised if you're struggling to meet 2.0 in the next couple years. Don't worry, this average doesn't have units, so it's all good.

02-13-2011, 06:12 PM
OP, are you in University? Usually the marks on these assignments aren't the important part, they're about forcing you to learn the material so you can pass the exams.

Do yourself a huge favor and learn this stuff on your own while you've got the opportunity to sit down with it. It'll help you in the long run when it comes to actually passing the course.