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#5844

shaunata (16 hours ago) []
Do animals fear humans on the sole basis of bipedalism? If so, if there were more bipedal animals would animals be as fearful? Does this fear come from the fact that homo erectus/ sapiens transition is still a young species? If so, if humans were around for a long enough period of time will animals feel comfortable around us?
#5847 to #5844

platinumaltaria (1 hour ago) []
Most animals are not inherently afraid of us, they're afraid of loud noises and sudden movements. If you stand still you can get birds to land and stick around near you because they don't see any threat.
Also you seem to have some misconceptions about evolution, humans are not a "young species" anymore than any other animals. Furthermore animals typically don't live for millennia and observe evolutionary change, so from their perspective things have always been this way; humans are the only animals that know there was ever any different life before ourselves.
Also you seem to have some misconceptions about evolution, humans are not a "young species" anymore than any other animals. Furthermore animals typically don't live for millennia and observe evolutionary change, so from their perspective things have always been this way; humans are the only animals that know there was ever any different life before ourselves.
#5840

heartlessrobot (20 hours ago) []
Wasn't sure where to ask so I figured here would be fine
I took 8 500mg tylenol 3 hours ago, and just took 12 more.
My headache hasn't gone away. These headaches pop up every two or three days. Sometimes I get a week's reprieve, but that's rare.
I've been drinking water, eating a reasonable amount, even tried essential oils headache cures.
Nothing.
Nothing but pain.
It makes me want to take a fucking powerdrill to my skull.
Any advice? I should probably see a doctor, but having "mind numbing headaches" will probably disqualify me from military service.
I took 8 500mg tylenol 3 hours ago, and just took 12 more.
My headache hasn't gone away. These headaches pop up every two or three days. Sometimes I get a week's reprieve, but that's rare.
I've been drinking water, eating a reasonable amount, even tried essential oils headache cures.
Nothing.
Nothing but pain.
It makes me want to take a fucking powerdrill to my skull.
Any advice? I should probably see a doctor, but having "mind numbing headaches" will probably disqualify me from military service.
#5841 to #5840

mublerking (19 hours ago) []
Being dead also disqualifies you from military service...so maybe go see a medical professional. You very well could be poisoning yourself considering 4000mg is an overdose, and thats over the course of a day not all at once. Tylenol isn't a magic cure all, it only affects headaches caused by certain things and taking more doesn't help.
#5846 to #5841

heartlessrobot (7 hours ago) []
I mean, it didn't kill me, it's been quite a while, and I feel fine except for gross morning breath. So, maybe just cut back on the tylenol?
#5839

masterreposter (23 hours ago) []
We should start calling everything by it's scientific name
#5842 to #5839

platinumaltaria (19 hours ago) []
That would be pointless, everyone knows what aspirin is; relatively few people know it's composition.
#5843 to #5842

masterreposter (18 hours ago) []
i think it would be funny for a while at least to screw with some people
#5833

masterreposter (10/22/2016) []
If my doctor gives me fat burning pills for weight loss, will I get excess skin? If so, will it go away if the person's body is adaptable and good at healing?
#5835 to #5834

masterreposter (10/23/2016) []
I do. I was just wondering if your skin adapted as fast to it since people say you need skin removed with lipo
#5826

naafi (10/20/2016) []
Why is porn so loud but everything else is too quiet? when I am watching normal videos I need to put my volume to 70% just to hear it but when I am watching porn I have the volume on 20% and it's still to loud
#5829 to #5826

platinumaltaria (10/20/2016) []
Video volume is entirely arbitrary, different sites will set the base volume differently.
#5832 to #5827

mublerking (10/20/2016) []
Its probably mostly due to confirmation bias (its not neccessarily true, you just remember it beig true because the times it wasnt true dont stick out). It couldnalso be explained by normal videos the audio being edited because they want to have a higher quality to them, whereas porn its probably just loosely thrown together and the audio isnt great.
#5828 to #5825

platinumaltaria (10/20/2016) []
Hair changes colour when the follicles stop producing pigment, that goes for all colours of hair. Hair can become either grey or white, and it depends on the structure of the hair itself rather than any pigment.
#5823

hirollin (10/17/2016) []
Not quite sure if this qualifies as science, more of math.
y>4
Quantity A
3y+2
____
5
Quanitity B
Y
Question is which is bigger. Answer is B, but I have no idea why.
y>4
Quantity A
3y+2
____
5
Quanitity B
Y
Question is which is bigger. Answer is B, but I have no idea why.
#5824 to #5823

mublerking (10/17/2016) []
these two things are basically two functions:
y1=(3x+2)/5
y2=x
Now, these curves meet at x=1. Theres three cases then:
for x<1 we know that y1>y2
for x=1 we know that y1=y2
for x>1 we know that y1<y2
This problem essentially asks which is greater for x>4. Well x>4 also satisfies x>1, thus
y1<y2, or the second one is greater.
If you wanted to solve this yourself a good way to view this in the future is to (if you're just talking about lines like you are here) draw out the graphs of the two lines and find where they intersect. This will help you visualize which one is greater and in which regions.
y1=(3x+2)/5
y2=x
Now, these curves meet at x=1. Theres three cases then:
for x<1 we know that y1>y2
for x=1 we know that y1=y2
for x>1 we know that y1<y2
This problem essentially asks which is greater for x>4. Well x>4 also satisfies x>1, thus
y1<y2, or the second one is greater.
If you wanted to solve this yourself a good way to view this in the future is to (if you're just talking about lines like you are here) draw out the graphs of the two lines and find where they intersect. This will help you visualize which one is greater and in which regions.
#5809

warrenzthehero (10/12/2016) []
I'm doing some Astro homework and I've come across a question that seems... too easy, and I'd like some help figuring out if I'm looking at this question right.
The question, paraphrased, asks:
As a Red Giant, our Sun will reach a peak Luminosity of 3000LSun. So the Earth would need to absorb 3,000 times as much energy as it does now. Therefore, it will need o radiate 3,000 times as much thermal energy to maintain it s surface temperature (assumned to be 300K). Estimate the temperature the Earth's surface will need to be to radiate that much energy.
At first glance this seems like a simple multiplication problem, but that seems too easy. Thoughts?
The question, paraphrased, asks:
As a Red Giant, our Sun will reach a peak Luminosity of 3000LSun. So the Earth would need to absorb 3,000 times as much energy as it does now. Therefore, it will need o radiate 3,000 times as much thermal energy to maintain it s surface temperature (assumned to be 300K). Estimate the temperature the Earth's surface will need to be to radiate that much energy.
At first glance this seems like a simple multiplication problem, but that seems too easy. Thoughts?
#5822 to #5809

drastronomy (10/13/2016) []
You have to use the stefan boltzmann law and assume earth is a black body
i think
i think
#5810 to #5809

mublerking (10/12/2016) []
You should try using the boltzman law for radiative cooling and equating the dE/dt to 3000 times the current dE/dt (using the 300K and the temperature of space.)
hyperphysics.phyastr.gsu.edu/Hbase/thermo/cootime.html
hyperphysics.phyastr.gsu.edu/Hbase/thermo/cootime.html
#5811 to #5810

warrenzthehero (10/12/2016) []
So use 300K to calculate Earth's current L, multiply that by 3,000, and then solve out the new T?
#5812 to #5811

mublerking (10/12/2016) [] Yeah, you can include the temp of space but i didnt since its negligible.
(Ive hidden it in a spoiler just in case you want to do it on your own first)
Click to show spoiler
(Ive hidden it in a spoiler just in case you want to do it on your own first)
#5813 to #5812

warrenzthehero (10/12/2016) [] Holy damn it worked. Thanks man. I din't use Radiative Cooling Time but instead used the L = 4(pi)R^2(sigma)T^4 and arrived at the same result. It was just the logical step I was missing.
I don't have much on this computer, but take this Chandra wallpaper as thanks, friend.
Click to show spoiler
I don't have much on this computer, but take this Chandra wallpaper as thanks, friend.
#5814 to #5813

mublerking (10/12/2016) []
Theyre essentially the same, yours just has the area of a sphere already in and removes the efficiency coefficient (likely because the atmosphere to space isnt a particularly lossy barrier), but anyways thats beside the point.
And no problem man, you got anymore physics questions you know where to find me.
And no problem man, you got anymore physics questions you know where to find me.
#5815 to #5814

warrenzthehero (10/13/2016) []
Alright friend, I've got a new question. Mostly I'm having difficulty figuring out which formula to use here. I'm given the distance in lightyears and the Luminosity of Sirius (8.6ly and 26LSun) and that of a hypothetical Betelgeuse supernova (643ly and 1x10^10 LSun and I'm asked to find out how much brighter the supernova would be than Sirius.
It seems I'd have to find their magnitudes and compare them, which means I'd have to use Flux. Problem is, I'm not certain what to use for r in F= L / 4(Pi)r^2. Is that lightyears? meters? AU?
Once I do that, I'm still not certain what to do. Would I use m = 2.5 log(Flux / FluxVega) and compare the resulting Magnitudes?
It seems I'd have to find their magnitudes and compare them, which means I'd have to use Flux. Problem is, I'm not certain what to use for r in F= L / 4(Pi)r^2. Is that lightyears? meters? AU?
Once I do that, I'm still not certain what to do. Would I use m = 2.5 log(Flux / FluxVega) and compare the resulting Magnitudes?
#5816 to #5815

mublerking (10/13/2016) []
You're using the correct equation (you do want flux. since it tells you how much light is taken in per area). In this case the units don't particularly matter because its looking for a ratio, so a dimensionless quantity, the units should cancel anyways (and I showed that).
In general it shouldn't matter which units you use as long as you're consistent (e.g. if they instead asked you how much light collided with earth, if you used r in lightyears to calculate surface area of the light sphere, you'd then need to use lightyears for radius of earth to calculate its cross sectional area)
In general it shouldn't matter which units you use as long as you're consistent (e.g. if they instead asked you how much light collided with earth, if you used r in lightyears to calculate surface area of the light sphere, you'd then need to use lightyears for radius of earth to calculate its cross sectional area)
#5818 to #5816

warrenzthehero (10/13/2016) []
Alright this is driving me up the goddamn wall.
I'm tasked with finding the mass of a hypothetical Black Hole whose denisty is that of water (1g/cm^3). I've worked through the process several times and even googled the answer and none of it works. My calculation is also waaaay off from any of the Google answers.
You need
V=4/3 (Pi) r^3
D = M/V
r=2GM/c^2
right?
So you start with
1 = 3M / [4(Pi)(2GM/c^2)^3]
Which becomes
1 = 3M / [32 (Pi) G^3 M^3 (1/c^2)^3]
1= 3Mc^6 / 32 (Pi) G^3 M^3
1= 3c^6 / 32 (Pi) G^3 M^2
And that is a LOT of constants. We'll call that whole mess K for a moment
1 = K (1/M^2) So,
M^2 = K, or
M = sqrt(K)
But when I run that through my calculator, I get 8.486x10^39 which is waaaaay off target. Other sources put it somewhere near 10^9, give or take a factor of 10 (all of those sources' answers have also not worked).
I just don't get what I'm doing wrong here.
I'm tasked with finding the mass of a hypothetical Black Hole whose denisty is that of water (1g/cm^3). I've worked through the process several times and even googled the answer and none of it works. My calculation is also waaaay off from any of the Google answers.
You need
V=4/3 (Pi) r^3
D = M/V
r=2GM/c^2
right?
So you start with
1 = 3M / [4(Pi)(2GM/c^2)^3]
Which becomes
1 = 3M / [32 (Pi) G^3 M^3 (1/c^2)^3]
1= 3Mc^6 / 32 (Pi) G^3 M^3
1= 3c^6 / 32 (Pi) G^3 M^2
And that is a LOT of constants. We'll call that whole mess K for a moment
1 = K (1/M^2) So,
M^2 = K, or
M = sqrt(K)
But when I run that through my calculator, I get 8.486x10^39 which is waaaaay off target. Other sources put it somewhere near 10^9, give or take a factor of 10 (all of those sources' answers have also not worked).
I just don't get what I'm doing wrong here.
#5819 to #5818

mublerking (10/13/2016) []
I mean, your calculation seems correct (other than the density which should be converted to kg/m^3 to keep units consistent). The mass of our sun is about 1.9*10^30, this would put this black hole at a mass of about 10^8 solar masses, or 100million solar masses. this doesn't seem too unreasonable, as this is the typical mass of supermassive blackholes which are well known as having densities similar to that of water (if you believe in the relativistic interpretation as opposed to the hologram interpretation which makes density kind of irrelevant, but I digress)
Another less rigorous equation for density of black hole is:
(1.8x10^16 g/cm3) x (Msun / M)^2, with a mass M=10^8Msun this would make it about 1.8, so this final result does actually seem reasonable.
Another less rigorous equation for density of black hole is:
(1.8x10^16 g/cm3) x (Msun / M)^2, with a mass M=10^8Msun this would make it about 1.8, so this final result does actually seem reasonable.
#5820 to #5819

warrenzthehero (10/13/2016) []
I tried one last calculation before you responded (the online homework uses an attempt system and this was my last) and got it wrong. The answer it said was correct was something like 1.35x10^10, I think.
#5821 to #5820

mublerking (10/13/2016) []
If you put it in solar masses as opposed to kg its 1.36*10^8, which is very close. (You get the same if you use the density equation). Thats in solar masses, the other answer may be wrong because its in kg.
#5817 to #5816

warrenzthehero (10/13/2016) [] You angel, you absolute marvel.
Click to show spoiler
#5798

masterreposter (10/07/2016) []
if you go back in time, why wouldnt the earth be in the spot that it was in during that time in space but you would?
#5802 to #5798

khankhan (10/07/2016) []
Interesting point, is there anything such as a fixed point in the universe? Moreso, if you traveled back in time would your body's particles compress due to the fact that they've experienced the expansion of the universe longer? Plenty of questions we don't know the answer to
#5799 to #5798

mublerking (10/07/2016) []
I always assumed the power to go back in time would be less just you shifting through time, and more just your mind is in your past self's body. Hence why you'd still be on earth and not in the middle of space.
#5800 to #5799

masterreposter (10/07/2016) []
theres a theory that if you went back in time than earth wouldnt physically be in the spot where you went back in time to because earth would still be moving forward in the same spot as if you were in the present. i dont get it either because i think of going to the physical earth thats part of that past, like how you expect people and things to be there too
#5801 to #5800

mublerking (10/07/2016) []
I mean that theory just gets way too complex, in that okay earth is in a new position but what about the sun, when you go back in time you and the entire earth stay in the same spot but does the sun? then you say okay fine the sun is in the new position too, but then what about the rest of the galaxy? okay fine that too, and eventually you'd have to conclude that everything would be the way it is exactly now just "in the past", which is a trivial distinction because "time" isn't an observable quantity (like energy. you can't measure your energy only differences in energy. So if we defined a new timeframe t'=t+c that only differs by a guage transform. well thats a trivial distinction because t'1t'2=t1+ct2+c=t1t2 so its the exact same.)
#5795

anon (10/06/2016) []
Two objects are in contact. A force is applied to object 1 and both objects accelerate. Object 1 is on wheels and object 2 is in contact with the ground. Find the constant for kinetic friction and draw two separate free body diagrams.
I can calculate the constant, however, I'm confused about the free body diagram. Specifically, the horizontal components for object 1. I assume that object 1 is frictionless since it's on wheels. So how does the kinetic friction from object 2 affect object 1?
Pic related, my free body diagram, I feel like object 1 is missing something.
I can calculate the constant, however, I'm confused about the free body diagram. Specifically, the horizontal components for object 1. I assume that object 1 is frictionless since it's on wheels. So how does the kinetic friction from object 2 affect object 1?
Pic related, my free body diagram, I feel like object 1 is missing something.
#5796 to #5795

Fgner (10/06/2016) []
It's actually fairly simple let's just hope my physics is up to scratch still ! When I do problems, I like to isolate the systems into lots of smaller chunks then solve with linear algebra, it makes it much easier to handle.
System 1
So, isolating the first (frictionless) block we see four forces acting upon it, and we know it's moving horizontally. As such, F1_y is equal to 0, the forces must balance. And F1_x must be equal to it's mass times acceleration (F=ma). And to make it easier to write, F_2on1 = F_t (transferred)
Ah shit, you already did that part. I'm a dumbass for not reading your entire comment first. Well... I guess you can check your work now.
Your Answer
Other than renaming F_(2 on 1) and F_(1 on 2), to be identical, you're golden.
System 1
So, isolating the first (frictionless) block we see four forces acting upon it, and we know it's moving horizontally. As such, F1_y is equal to 0, the forces must balance. And F1_x must be equal to it's mass times acceleration (F=ma). And to make it easier to write, F_2on1 = F_t (transferred)
+↑ Σ F1_y: F_1N  (m_1*g) = 0 +⟶ Σ F1_x: F  F_t = (m_1*a) [code] We don't need to solve or substitute with anything in F1_y, so we can just toss it out. The only unknown value in F1_x is that nasty F_t, so let's solve in terms of that. [code] F_t = F  (m_1 * a) [code] Sweet, first system down. [big]System 2[big] Again, we have four separate forces being applied: [code] +↑ Σ F2_y: F_2N  (m_2*g) = 0 +⟶ Σ F2_x: F_t  f_k = (m_2*a) [code] Now, of course, f_k can be subbed for (F_2N * u_k), and we can substitute F_t for what we found for it's value in the first system. [code]+⟶ Σ F2_x: F  (m_1 * a)  (F_2N * u_k) = (m_2*a) [code] And now we simplify for the value we desire. In this case, u_k: [code] u_k = (F  (m_2 * a + m_1 * a) / (m_2 * g) [code] [big]Fast Method[big] Or, more simply, using F=ma for the whole system: [code] F  f_k = (m_1 + m_2) * a F  (m * g * u_k) = (m_1* a + m_2*a) u_k = ...
Ah shit, you already did that part. I'm a dumbass for not reading your entire comment first. Well... I guess you can check your work now.
Your Answer
Other than renaming F_(2 on 1) and F_(1 on 2), to be identical, you're golden.
#5797 to #5796

Fgner (10/06/2016) []
Oh, Funnyjunk and your shitty formatting. Now I know never to use that code shit again, it didn't even make it monospaced...
You can ignore most of that comment, just read the bottom. Diagram = perfect other than naming. You may also want to include one more arrow to show acceleration.
You can ignore most of that comment, just read the bottom. Diagram = perfect other than naming. You may also want to include one more arrow to show acceleration.