The discussion focuses on solving a homework problem involving water pressure using the equations P2=P1+pgh. The initial attempt incorrectly suggests plugging in values directly without properly applying the formula. It clarifies that 1.3x10^5 pascals is the pressure in chamber 2, not chamber 1. To find the pressure in chamber 1, users are advised to rearrange the formula for pressure to calculate force using the known area. The conversation emphasizes the importance of correctly applying the given equations to arrive at the right solution.
#1
yesshh
3
0
Homework Statement
Homework Equations
P2=P1+pgh
The Attempt at a Solution
I thought you would just plug in the pressure in pascals to p1 and add [(1000 kg/m3)(9.8m/s2)(sqrt 2)]?
Okay find out the pressure in chamber 1. Pressure=force/area.
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#3
yesshh
3
0
So the answer is just 1.3x10^5 divided by 2?
Why?
#4
hollidca
11
0
No, 1.3x10^5 is pressure in chamber 2. Use the forumla given to you in the question to calcualte the pressure in chamber 1. You know the area of chamber 1 and you have calculated the pressure for chamber 1 so rearrange the formula pressure=force/area to give you the force.
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...
Figure 1 Overall Structure Diagram
Figure 2: Top view of the piston when it is cylindrical
A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.
Figure 3: Modifying the structure to incorporate a fixed internal piston
When I modify the piston...
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...