Physical Science class question
122694
Most of the water molecules in the air in your freezer have sublimated to a solid state inside. Therefore, there are less water molecules in the air making it less 'humid'.
215968
oh, wow, this is a physics question.\n\nto start, we remember E = mc^2\nthe problem gives 4 grams matter and 2 grams antimatter.\nmy solution would be that 2 grams of matter and 2 grams of antimatter annihilate to produce 4c^2 (2 + 2 = 4) energy.\nthe excess 2 grams of matter remains unaffected.\nthe answer is D.
528811
You're in luck.  I teach physics and I know exactly how to answer this question (it's based on the work-energy principle).  It's pretty simple, really.  But it would be best if we discussed it by messenger.  Yahoo Messenger me, my ID is\nfortitudinousskeptic\n\n- Kevin
471898
If m1 is the mass of the moving particle and m2 is the mass of the particle at rest, then \n\nm2 / (m1+m2) of the kinetic energy of the moving particle will be lost after inelastic collision.\n\n\nBy law of conservation of momentum the final velocity of the two together is \n\nV = m1 v1/ (m1 +m2) becasuse v2=0.\n\nAfter collision the kinetice energy is 0.5 (m1+m2) V^2.\n\nBefore collision the kinetic energy was only 0.5 m1 v^2.\n\nUsing these relations we get the above result.
491970
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
516029
plz ask in physics category
230398
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
195405
i doubt your teacher couldnt answer these unless (s)he is not actually a physics teacher\n\nc and a\n\nthe initial momentum is 4kgm/s\nthe final momentum is conserved so it is still 4 kgm/s\nmomentum is mass*velocity\nmass is 1+3= 4kg\nv = p/m = 4kgm/s / 4kg = 1 m/s\n\nentropy is randomness or disorder, when the rooms are different temps, its like the temps are "sorted". there is order to the temperature. when they combine, there is no more order, the temperatures are more random instead of being focused around 2 different temps so disorder, or entropy increases
367681
(a) Work done by gravity = change in potential energy of the book. Now change in potential energy = m*g*h2 - m*g*h1=2*10*(10-1.5) =  170J. So, work done by gravity is 170J\n(b)Change in gravitational potential energy = work done = 170J as explained above.\n(c) Potential energy when textbook is released = m*g*h1 = 2*10*10=200J\n(d) Potential energy when it reaches the hands = m*g*h2 = 2*10*1.5=30J\n(e) If U =100J at ground level, work done by gravity = 170J, the work done does not change with change in reference.\n(f) change in potential energy also remains 170J\n(g) U at release point = 100 + 2*10*10 = 300J\n(h) U at hands = 100 + 2*10*1.5 = 130J
523448
this question is sooo odd. there is no way someone will no this without actually doing an experiment. you dont deserve to pass the class if you didnt do an experiment to find it.
457207
My magic 8 ball says no. \n\nSo does my physics professor.
59130
Hey I'm studying physical and chemical equilibrium too, currently.The answer is that it will definitely have units when the concentrations have different powers.Just try and work it out with the units and it will be very clear.
321116
I am sorry I can not provide you with an answer to your question. However it might help you get more answers if you put this question in again but put it in the Science and Mathmatics section. Under Physics. I think there are more people willing to help you out there!
353556
What teacher still gives exercises in such mixed units? Degrees Fahrenheit and Kelvins. Distance in feet....but the results is required in grams?\n\nThe volume of air is 16000 ft^3 or 453 m^3. The density of air is 1.293 gram/liter (I looked that up) or 1293 gram/m^3. We have therefore 5.86 x 10^5 grams of air to heat.\nThe temperature change is 30 deg F or 16.7 deg K.\nSo the total heat needed is 1.01 x 16.7 x 5.86 x 10^5 -9.88 x 10^7 Joule. \nThe combustion of methane provides 890.8 x 10^3 J/mol, which means that we need 111 mol of methane or 111 x (12+4x1) = 1775 gram of CH4\n\nThe reaction equation is CH4 + 2 O2 -> CO2 + 2 H2O.\nSo one mol of methane will produce one mol of CO2. A total of 111 mol are produced, or 111 x (12+2x16) = 4884 gram of CO2.\n\n(And all that stuff goes in the atmosphere warming up the greenhouse a little more.)
156605
The answer is NOT pressure... The correct answer is "NO"\n(That's why kindergarten students can answer it.)
428609
57) P1V1/T1 = P2V2/T2\n(temp in kelvins)\n140 x 8.00 / 293 = 1 x V2 / 273\nV2= 1043.55 L but sig figs is 3 so 1040 L\n\n93) uhh I'm just learning this myself.. very confused.. something about london dispersion, hydrogen bonding, dipole-dipole forces? perhaps someone else can take this part... and give them the 10 points I don't really care
553113
Are the containers the same mass? \nAre the contents of the container the same material?\n\nIf the answer to both questions is 'Yes' 1st, the motion of the molecules is the same but, 2nd, with the 200g there is twice as much energy in the water contents as you have twice as much .