|
To any physics people out there:
1) The sole of a certain tennis shoe has a shear modulus of 4x10^7. If the height of the sole is doubled, the strain will:
A. decrease by a factor of two
B. remain the same
C. increase by a factor of two
D. increase by a factor of four
2) A brick w/ a density of 1.4x10^3 kg/m^3 is placed on top of a piece of styrofoam floating on water. If one half the volume of the styrofoam sinks below the water, what is the ratio of the volume of the styrofoam compared to the volume of the brick? (Assume the styrofoam is massless)
A. 0.7
B. 1.4
C. 2.8
D. 5.6
I know the answers to both of the questions, I just understand the reasoning on how to obtain the answers. If someone can explain why the answer is what it is, I would greatly appreciate it. Thanks in advance.
Answers: 1. B 2. C
 
|
Micronesia (If he didn't go to sleep yet) to the rescue!
|
T_T
i fail once again at physics @ TL.net
|
Looking at number one, it's mainly just trying to scare you with shear modulus.
The equation for shear modulus is, SM = stress (F / A) / strain (delta H / totalH) -- SM is a constant as is stress (same person in the shoes, only changing the height of sole). Obviously, then, strain is going to be constant. So it'll remain the same.
|
United States24773 Posts
I solved 2. Let me edit with my explanation.
Edit:
2) A brick w/ a density of 1.4x10^3 kg/m^3 is placed on top of a piece of styrofoam floating on water. If one half the volume of the styrofoam sinks below the water, what is the ratio of the volume of the styrofoam compared to the volume of the brick? (Assume the styrofoam is massless)
The rule with floating objects is that the weight of the object floating in the water is equal to the weight of the water it displaces (if it needs to displace water greater than its volume equivalent in order for that to be true, it will just sink). The weight of the brick is going to be equal to the weight of water that occupies half the volume of the Styrofoam (the problem says half the Styrofoam is submerged).
Density of brick = 1.4x10^3 kg/m^3
Density of water = 1x10^3 kg/m^3
Volume of Brick = Vb
Volume of Styrofoam = Vs
Volume of water displaced = Vs/2
1) Weight of brick = Weight of Water Displaced
2) Weight of water displaced = density_water * Vs/2
3) Weight of brick = density_brick * Vb
Set 2 and 3 equal, as per 1:
d_w * Vs/2 = d_b * Vb
Vs/Vb = 2*d_w/d_b = 2.8
|
(01:10:24) (+micronesia) SayaSP thanks for electing me to do physics lol
Well, you ARE a freakin' physics teacher.
|
United States24773 Posts
On July 20 2008 14:10 SayaSP wrote:
(01:10:24) (+micronesia) SayaSP thanks for electing me to do physics lol
Well, you ARE a freakin' physics teacher.
Where's my money then.
|
On July 20 2008 13:17 GrayArea wrote:
To any physics people out there:
1) The sole of a certain tennis shoe has a shear modulus of 4x10^7. If the height of the sole is doubled, the strain will:
A. decrease by a factor of two
B. remain the same
C. increase by a factor of two
D. increase by a factor of four
2) A brick w/ a density of 1.4x10^3 kg/m^3 is placed on top of a piece of styrofoam floating on water. If one half the volume of the styrofoam sinks below the water, what is the ratio of the volume of the styrofoam compared to the volume of the brick? (Assume the styrofoam is massless)
A. 0.7
B. 1.4
C. 2.8
D. 5.6
I know the answers to both of the questions, I just understand the reasoning on how to obtain the answers. If someone can explain why the answer is what it is, I would greatly appreciate it. Thanks in advance.
Answers: 1. B 2. C
Well number 1) is just about understanding the meaning of a shear modulus. I remember shear modulus as the "friction resistance" of a material. It's essentially how strong it resists a plane moving against it, whilst it's being pushed against that plane. The thickness of the sole is clearly not going to affect this strength.
Number 2) hmm this is a bit tricky.
Always remember Archimedes principle whenever you have things in water.
The "upthrust" force is equal to the weight of the water displaced.
Water conveniently has a density of 1kg/L, which helps the calculations a lot.
So let the volume of the brick be B and the volume of the Styrofoam be F.
1.4* B = 0.5(since it's half the foam)* F * 1(the density of the water displaced)
1.4 * B = 0.5 * F
F = 2.8B as stated.
Edit: Shit, by the time i'd typed it out people beat me too it :p
|
1 I have no clue, probably some formula u can use...
2 I'll have a go. The only thing I knew is
dispaced water mass = mass lifted.
so with that in mind...
2) A brick w/ a density of D is placed on top of a piece of styrofoam floating on water. If one half the volume of the styrofoam sinks below the water, what is the ratio of the volume of the styrofoam compared to the volume of the brick? (Assume the styrofoam is massless)
Let's imagine brick w/ density D
Let's imagine water w/ density K
Let's imagine brick w/ volume V
Let's imagine foam w/ volume W
Brick mass = Dispaced Water mass
what it means?
Brick with density D, volume of V
Dispaced water with density K, volume of 1/2W
D*V = K*W/2
What do you want? Ratio of Brick vs Foam on volume, so V/W
so rearrange...
V/W = K/2/D
|
Oh yeah what physics is this? Go to Office hours often if it's university
:D
|
United States24773 Posts
No response from the OP?
D:
|
Sorry about the lack of response, I typed it out then signed off and went to sleep. Thanks so much for the responses you guys. The problems are from an MCAT physics study book, and it has explanations in the back except that some explanations are written kind of poorly such that I am not able to understand the reasoning behind it.
I was just unsure about the first one because I was trying to apply an equation to it. Pervect made a good response, thanks pervect. What I don't understand though is if the height of the sole is doubled, won't it affect the area? As in: S.M. = stress / strain = (F/A) / (delta h/h). So by increasing the height of the sole, the A will increase which will reduce the stress factor. Which will ultimately reduce the S.M.? But the S.M. is a constant so I am just getting confused on how the changing of the height is affecting either the stress or the strain.
The second question was explained very well by micronesia, thanks micronesia. That really makes sense when you apply all the formulas and show step by step how to arrive at the answer. The answer explanation just wrote some response that sounded kind of esoteric to a physics learning student. Thanks also to everyone else who replied, I really appreciate it. 
|
United States24773 Posts
Excellent. Just be careful when reading over an explanation like the one I provided. What used to happen to me occasionally was, someone would explain something to me, it would seem clear as day, I'd be confident, but then I wouldn't be able to do it on the test XD. You might want to try the same, or a similar problem again later without looking at this, just to see if you really have it yet. Good luck with your MCAT.
|
Ya, I still have till next spring until I take the exam, I am just trying to study as much as I can before then. I will be practicing a lot and doing a lot of problems to get this stuff down, especially because physics is my weaker of subjects. Hopefully by test time, I will be doing these problems in my sleep. 
|
I was just unsure about the first one because I was trying to apply an equation to it. Pervect made a good response, thanks pervect. What I don't understand though is if the height of the sole is doubled, won't it affect the area? As in: S.M. = stress / strain = (F/A) / (delta h/h). So by increasing the height of the sole, the A will increase which will reduce the stress factor. Which will ultimately reduce the S.M.? But the S.M. is a constant so I am just getting confused on how the changing of the height is affecting either the stress or the strain.
No, have a look at the picture on this page:
http://en.wikipedia.org/wiki/Shear_modulus
The area is the area of the object tangential to the direction of the force. The height of the object is independent of the area. The height and area sort of specify the characteristics of the object, while F and deltah tells you the object's response to a shearing force.
|
On July 21 2008 10:05 skyglow1 wrote:Show nested quote +I was just unsure about the first one because I was trying to apply an equation to it. Pervect made a good response, thanks pervect. What I don't understand though is if the height of the sole is doubled, won't it affect the area? As in: S.M. = stress / strain = (F/A) / (delta h/h). So by increasing the height of the sole, the A will increase which will reduce the stress factor. Which will ultimately reduce the S.M.? But the S.M. is a constant so I am just getting confused on how the changing of the height is affecting either the stress or the strain. No, have a look at the picture on this page: http://en.wikipedia.org/wiki/Shear_modulusThe area is the area of the object tangential to the direction of the force. The height of the object is independent of the area.
Ohhhhhhhhhhh, thanks!! that makes it so much more clear! You always make nice posts in my blog, its like you know exactly what I am thinking in all my posts, especially that one o-chem post where I got an A-. I just want to say I appreciate your omnipotence, thanks man. 
|
Whoopsie I was editing some more stuff when you replied but I guess you don't need it lol.
No problem :D
|
micro when i have mcats im going to need you to do my fizzix also
|
United States24773 Posts
On July 21 2008 11:21 Caller wrote:
micro when i have mcats im going to need you to do my fizzix also
Not after the way you brutally 2v1 rushed me today while my partner took four hours getting his army to my base.
|
|
|
|
|
|
EPT
