Nicholas Kang

Ok, though I don't take biology now after high school. I still remember what I learn. This is more of a biochemistry question. The explanation is simple. First, you need to understand what is osmosis, osmotic pressure and how osmoregulation works. Osmoregulation (See:https://en.wikipedia.org/wiki/Osmoregulation) is basically the regulation of an organism's body water content to maintain its osmotic pressure. You also need to understand the term endoosmosis, exosmosis, hypertocnic and hypotonic etc. Ok ,we start with the simple ones first. 3 types of solutions exist: solutions hypotonic to cell, solutions hypertonic to cell or solutions isotonic to cell. Say you place an animal cell into a solution that is hypotonic to the cell. The hypotonic solution means that is has less solute in the solution, thus less concentrated and has lower osmotic pressure. So, relatively, the cell has higher osmotic pressure, and water flows across the cell membrane from the outside (solution) to the intracellular space (inside the cell). The cell will accumulate water over time and swells and may eventually burst. The opposite happens to the animal cell when it is placed in a hypertonic solution. It has high solute content and thus more concentrated. So, an animal cell immersed in it tends to lose water to the surrounding, and the animal cell shrinks in size. For an erythrocyte (Red blood cell), it is said to crenate. For plant cell, the condition differs a bit, given that plant cells have cell walls, which is absent in animal cells. Plant wall gives the plant cell strength. So, when a plant cell is immersed in a hypotonic solution, though water enters the plant cell, the plant cell would not burst, but remain turgid, i.e. the vacuoles are large and the plant cell is stiff. However, if immersed in a hypertonic solution, the plant cell will lose water like an animal cell, thus shrinks in size, and we say the plan wilts. If the dry condition is prolonged, it leads to the death of plant. An isotonic solution is the solution which has the same solute concentration as that of the organism's body. Hence, it is the optimum condition for an organism to live in and the cells won't lose or gain any water, and hence no change in the organism's body size is observed. Here is a website that teaches you the 3 concepts. http://www.phschool.com/science/biology_place/biocoach/biomembrane1/solutions.html --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- So, given such extremes condition, organism (composed of cells) learn to adapt themselves to their surrounding environment. In you question, saltwater fish is the organism in question. Saltwater fish lives in a hypertonic environment, that is their surrounding fluid (sea water) has high salt content, thus hypertonic to the cells in saltwater fish. If no preventive measure is taken, the cells will loss water to its surrounding quickly and they will die due to lack of water. So, one way to overcome this is to increase the organism's body salinity to make them even more concentrated to the surrounding environment (seawater in this situation, or any salty aquatic environment). Now, it is their bodies that are more hypertonic than the surrounding salty water, thus instead of losing water to the surrounding, they gain water from the surrounding. So, the answer to your question is A. Good luck with your studies!

Ok, I think I need to further clarify my case. I don't mean someone just sits on lunar surface and watch Earth wobbling. I mean, get yourself up and high away from lunar surface and stay idle there, see how Earth moves around the Moon. In my test, it does "orbit" around the Moon this way (Moon is observer, Sun is focus, hence Earth is moving): Then, Earth to the right And finally it is between moon and Sun Closer view: Earth is missing which proves that it is really between Moon and Sun. Then, Earth to the left And finally back again (Moon between Earth and Sun). Earth completes one "orbit" around the Moon. Earth does not cover the Moon because Moon is not coplanar with the ecliptic. (That's why you don't have solar and lunar eclipse every months. They only happen when they are collinear.) Here is the orbit plot diagram that I mentioned. My friend told me the white line is Earth's orbit plot. So, I want to know whether the orbit plot diagram is correct or not. In my case, I want to get from low lunar orbit to Earth-moon Lagrange point 1, so the yellow circle is the Moon and the green line is my planned trajectory, but I wonder what the white line is. My friend said that is Earth's trajectory as seem from Moon. So I want to know if that is really the case. And if he is correct, what is the orbital period of Earth around the Moon as seen from the Moon then? Thanks for answering. Nicholas.

Hello I am working on an orbit trajectory plot tool with my friend from the US. I have some question to ask all of you about this question. In our trajectory plot, we would like to fix Moon as the center of our orbit plot. Then, from our perspective (the moon), what would the Earth's orbit be or look like? From the moon's perspective, would the Earth seems like orbiting the Moon? If this is the case, how long does it take for Earth to "orbit" the Moon? The same period as lunar's orbit, i.e. 27 days? Thanks for answering. Nicholas.

Ok, Sensei. In the diagram, Cl atom is larger than B atom. Now, BabcockHall says he thinks Chlorine's orbitals will not (unsure) overlap effectively with the orbitals of boron. So, what do you think, Sensei? What is your opinion? (Simply put, what is your answer to my question? )

Well, this question is basically about mechanics, work and energy. There is no need to use suvat in this case. Strategy: Height already given-Hint: gravitational potential energy Speed given-Hint: kinetic energy Ok, apply principle of conservation of energy at both A and B Assume you already know the basic knowledge: EK = 0.5mv2 , EP = mgh, Assume g=10 ms-2 and let Fc = Frictional force, then EK at A + EP at A = EK at B + EP at B + Word done against friction 3(10)(6) + 0.5(3)(4)2 = (3)(10)(2) + 0.5(3)(8)2 + Fc(12) So, solve for Fc , you get Fc = 4N That is the frictional force is 4N. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ Now, again using the same concept, we can find how far will the block be when it stops, hint: final energy= Energy required to counteract frictional force. Ok, Total energy at B= total energy at A + work done against friction from A to B = total energy done to do work against friction when the block has stopped 3(10)(6) + 0.5(3)(4)2 = 4(Distance) Total distance traveled by the block until it stops = 204/4 = 51 meter But this distance include the distance from A to B, which is 12 m. So, distance traveled from point B to the point where it has stopped is (51-12)m or 39 m. OR If you choose to start from point B, then (3)(10)(2) + 0.5(3)(8)2 = 4(Distance) Distance = (60 + 96)/4 =39 m, which is the same answer as above. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ Remember, more practice is essential to develop an instinct that helps you solve Physics problems rapidly. But make sure you master the theories first before you begin. Good luck with you studies!

Are you sure? I thought Boron is larger than Chlorine. The atomic size of elements decreases across a period. And I have checked the atomic radius for both Boron and Chlorine. Boron: 87 pm Chlorine: 79 pm So, apparently, Boron is larger than Chlorine.

Can you explain it in detail? How does electronegativity affects hybridization?

Ok, finally someone replied. Thanks fiveworlds. Personally, I think this configuration is more stable, since hybridisation stabilizes a molecule. Apparently, if achieving octet or duplet is the only thing required for the formation of a particular molecule, why would you hybridise the central atom as well? So, it is clear that you hybridise atoms of molecules to make them stable and hybridise all atoms would definitely make the whole molecule stable. Am I wrong? Nicholas.

I have been offline for almost 2 years and now I am back. I hope you guys still remember me... Here is a question on physical chemistry. Simple question. It is about Valance Bond theory and the use of hybridisation/hybridization (depending on whether your English is American/British variant) concept. My teacher taught me how to explain Boron trichloride's structure using VB theory. Boron atom undergoes sp2 hybridisation. Each sp2 hybrid orbitals will have an unpaired electron that will be shared with the unpaired electron in Chlorine's 3p orbital. So, the orbital overlap diagram looks like this. (Check the attached file.) The 3 red orbitals are the 3 sp2 hybrid orbitals while the 3 green orbitals are from Chlorine's 3p orbital. My question is why Chlorine does not undergo hybridis(z)ation as well? Supposedly, each Chlorine atom should undergo sp3 hybridis(z)ation to produce 4 sp3 hybrid orbitals, one of which is shared with the sp2 hybrid orbital of Boron atom. So, 3 sp3 hybrid orbitals from 3 Chlorine atoms share 6 electrons with 3 sp2 hybrid orbitals from Boron atom. Thanks for answering. Glad to see you guys again. Regards, Nicholas.

Here is another one. Falcon 9 carrying OG2 in its fairing poised for launched at KSC. Thanks for enjoying the screenshot. Regards, Nicholas.

Hello, I have not been around SFN for quite some time. I was focusing on my studies and also flying to different orbits using Orbiter Space Flight Simulator. Nice to see all of you again. Here I would like to start a new thread for those who use Orbiter and love to share their beautiful screenshots. I will start first. Here are 2 (beautiful, or maybe opposite; beauty is in the eye of the beholder) screenshots captured using Orbiter. The first is ThaiCom 8 launched to elliptical transfer orbit after SECO of Falcon 9. The second is GAIA of ESA in Largrange Point 2 or L2 after launched from Kourou Spaceport aboard a Soyuz rocket. The black disc covering the Sun in the GAIA mission screenshot is Earth. Hopefully you enjoy the view. Regards, Nicholas.

I shall give it a try. Let's deal with the equations first. You had followed what studiot said. So, solve the three equations. (In Malaysia, we are taught that general quadratic equation is ax2+bx+c=0; we don't use z in place of c) 4a-2b+c=0--EQ1 16a+4b+c=0--EQ2 9a+3b+c=0--EQ3 EQ1 x2-- 8a-4b+2c=0 4b=8a+2c Substitute 4b=8a+2c into EQ1 You get: 16a+8a+2c=0 24a+3c=0 3c=-24a c=-8a Substitute c=-8a into EQ1 You get: 4a-2b-8a=0 -2b=-4a b=-2a Substitute b=-2a and c=-8a into EQ3 9a+3(-2a)+(-8a)=-15 9a-6a-8a=-15 -3a=-15 a=-3 Substitute a=-3 into b=-2a and c=-8a You get a=-3, b=-6, c=-24 Quadratic equation is 3x2-6x-24=0 To get the minimum value, I invoke the method of completing the square (This was taught in Form 4, 16-year-old syllabus in Malaysia) To complete the square, first you have to understand how it works. Let's say we have quadratic equations: (x-1)(x-1)=0 (x-2)(x-2)=0 (x-3)(x-3)=0 So, x2-2x+1=0 x2-4x+4=0 x2-6x+9=0 Notice the pattern of the above 3 equations. You can conclude: x2-2x=(x-1)2-12 x2-4x=(x-2)2-22 x2-6x=(x-3)2-32 So, x2-nx=(x-n/2)2-(n/2)2 Now, 3x2-6x-24=0 3(x2-2x)-24=0 3[(x-1)2-12]-24=0 3(x-1)2-3-24=0 3(x-1)2-27=0 After completing the square, you will obtain a formula that looks like a(x-s)2+v=0, where s represents the axis of symmetry and v represents the minimum/maximum value of the function/equation In this case x-1=0 thus axis of symmetry is x=1 whereas -27 is the minimum value of the equation. The answer is -27. (I posted a link in case you need some references. The link below directs you to a Malaysian Form 4/16-year-old Malaysia Education Certificate Syllabus-based Additional Mathematics website. Look for Chapter 2(quadratic equations) and 3(quadratic functions) for details.) http://spmaddmaths.onlinetuition.com.my/

studiot, I can roughly figure out where are you leading me to. Also, based on your last assumptions and what pavelcherepan had said, you must be talking be about thermal equilibrium with the surroundings. So, the final temperature must be equal to the temperature of the surroundings instead of just 10 degree Celsius. Is this what you mean? If so, this question must not be talking about a closed system, it must be an open system if you talk about surroundings.

Assumptions based on studiot. Good assumptions, studiot. At least you "fill in the gaps required." Solutions: Picture the problem Imagine an ice cube being placed in a glass of water. Ice cube is 0 degree Celsius. Water is 10 degree Celsius. We know that heat is transferred from the water to the ice so that thermal equilibrium is reached somewhere between 0 and 10 degree Celsius. So, we assume: heat absorbed by the ice = heat released by the water, that is, Qabsorbed = Qreleased We first assume that all sensible heat is removed from the water before latent heat of water sets in Qabsorbed Qreleased micelfusion mwatercwater(10-0) mice(334000 J/kg) 0.225(4180)(10-0) mice(334000 J/kg) 9405 J Now, we know the heat supplied when the water cools but not freeze at a constant temperature is 9405 J. Yet, some 334000 J of heat is needed to melt 1 kg of ice. So, we know the heat supplied supplied by the cooled water isn't enough. But thermal equilibrium is achieved since both the ice and water is 0 degree Celsius. The ice starts off at 0 degree Celsius and the water has cooled to 0 degree Celsius. No net heat flow occurs between 2 bodies. Thus, the temperature remains constant. The answer is 0 degree Celsius. You can check how "light the ice might be" by using the formula Q=ml, that is 9405=mice(334000) mice=9405/334000 =0.02816 kg, about 28.16 g of ice would be sufficient to absorb all 9405 J of heat and remains at a constant temperature with the cooled water. *I ignore the surrounding temperature in this case. Before you consider my answer, studiot, is my answer correct? I am a high school student and after reading the question I think I can answer it. Anyway, you should consult studiot first.

As a high school student still in the process of learning, I, Nicholas Kang would like to express my utmost gratitude to Dr. Tom Swanson for his meticulous teaching. Thank you Dr. Tom Swanson. You make my day. Anyway, I don' think the above post was a bad one. You see, I have studied painstakingly to comprehend Relativity, at least to some degree. (No, I am not asking for praise or anything in return, just a statement) Not all high school students can study modern physics and classical physics at the same time, especially when they need to go through the hardships of choosing either fields when answering high school exam questions. You would feel the pain of "betraying yourself" when writing wrong or "inaccurate" answers, as some might argue that classical physics isn't the big blunder of physics after all. Well, Dr. Swanson, is there any more errors in my post? Feel free to correct it. It's nice to have a teacher like you. Thank you. Your mentee, Nicholas Kang