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Help with Physics  Object submerged, displacement, density, pressure? 
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#1
Feb1812, 06:08 PM

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I have a few physics problems I need help with. Please show work or I wont know how to do it. My book has no example problems.
1. Suppose you balance a 5kg ball on the tip of your finger, which has an area of 1 cm^2. Show that the pressure on your finger is 49N/cm^2, which is 490kPa. For this question I get 500kPa, where is the 10kPa coming from? 2. 6kg piece of metal displaces 1 Liter of water when submerged. Show that the density is 6000kg/m^3 3. Deepest part of a lake has depth of 406m. Show that water pressure at this depth is 3978.8 kPa, and the total pressure there is 4080.1 kPa. 


#2
Feb1812, 11:24 PM

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We're not just going to give you fully worked out solutions. 


#3
Feb1812, 11:31 PM

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Yeah, well I dont even know where to begin. I can tell it is simple but I dont know what I need to do. If you honestly believe I'm just trying to have u do my home work, then use different numbers. I just want to know HOW to do these. Thanks.



#4
Feb1812, 11:59 PM

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Help with Physics  Object submerged, displacement, density, pressure?



#5
Feb1912, 12:00 AM

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For problem 2: What must the volume of the 1 L of displaced water be equal to?



#6
Feb1912, 12:04 AM

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For problem 3: what is the equation for the variation of pressure with depth?
THere is a reason why this site has a template for homework help posts that includes a complete statement of of the problem, a list of any *relevant equations*, and the work you've done so far. The second item, in particular gets you thinking about what physical relationships and principles are important to the problem. 


#7
Feb1912, 04:50 AM

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I second the stuff cepheid said. Also, for question 1, it looks like bpears has used g=10m/s^2 and the answer book used g=9.8m/s^2, so the difference in answers is only due to using different values for g.



#8
Feb1912, 02:06 PM

P: 10

Bruce, you are spot on. My instructor tells us to just use 10m/s^2 for gravity. Which I now see was a bad idea for our book. So, I get the 1st one. But the other two, I don't know how to do.
For number 2, I see that if it displaces 1L, then it must have the volume of 1L? And its safe to assume that the 1000 is multiplied by 6kg, but where does the L unit go? And what formula are you supposed to use on this? For number 3, I know P=pgh. so P=p(9.8m/s^2)(406m) Which without multiplying the density I get the 3978.8kPa. But I dont know why? What about density(p)? And then I don't know how to get total pressure of 4080.1kPa? 


#9
Feb1912, 05:18 PM

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For number 2, use the DEFINITION of density to compute the density.



#11
Feb1912, 07:17 PM

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are not getting 3978.8 kPa right now, with your calculation. You are getting 3978.8 m^{2}/s^{2} which is certainly not in units of pascals. You must pay attention to units. If you're wondering why multiplying by the density seemingly doesn't change the numerical value of the answer, then consider this: what is the density of water in SI units? So, what numerical value would you be multiplying by? What factor would you then divide the answer by to convert from Pa to kPa? As far as the total density goes you have to take into account not only the pressure due to the weight of the column of water above the lake bottom, but also the pressure due the weight of the column of *air* above that. In other words, the surface pressure is not 0. 


#12
Feb1912, 07:25 PM

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yes, I just read through bpears answer more carefully, and realised it didn't make sense. By wishful thinking, it looks like he jumped past the step of multiplying by density.



#13
Feb2012, 12:26 PM

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Ok, so for number 2, density=mass/volume. I found the volume of a Liter of water is 1m^3. So, then I get 6kg/m^3. How does the book get 6000kg/m^3?
And for 3, I got the 1st part but I don't know how to get total pressure? 


#14
Feb2012, 12:45 PM

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This should be obvious, if you've ever seen a onelitre carton of milk, then you know that it is not as big as a cubical box that measures one metre on each side! Then again, maybe product volumes are typically given in imperial units where you live, in which case you might not have a good intuition for metric ones. The thing I usually remember for converting this particular nonstandard unit (the litre) is that 1 mL = 1 cm^{3}. Also, anticipating future questions you might have: you have to be careful when converting units of volume. It can be tricky. For example, you might naively think that 1 m^{3} = 100 cm^{3}. This is FALSE. In fact: 1 m^{3} = (1 m)^{3} = (100 cm)^{3} = 100*100*100 cm^{3} = 1,000,000 cm^{3}. One cubic metre is a million cubic centimetres. That's an algebraic way to arrive at the answer. A geometric way to visualize this is that if you have a volume of 1 m^{3}, that means you have a cube that is one metre on each side, i.e. its dimensions are (1 m) x (1 m) x (1 m). So the dimensions of this cube in centimetres is 100 cm on each side, or (100 cm) x (100 cm) x (100 cm), and the volume in cubic centimetres is this product, which works out to 1,000,000 cm^{3}. In other words, if you have a small cube that is 1 cm on each side, then a million of these cubes, arranged in a 3D grid of 100 x 100 x 100, will fit inside a larger cube that is 1 m on each side. 


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