The normal body temperature of the certain mammal species may be modeled as a following equation: T(t) = 36.8-1.3sin((Pi/12)(t+2)), where t=0 corresponds to 12am (or 0:00), and t = 23 to 11pm (or 23:00) If the Earth suddenly started to rotate 3 times as fast as now, how would you change the equation in order to maintain the same values of the lowest and the highest temperatures? Also, the lowest and the highest temperatures must occur at the same time relative to the beginning and the end of the day as in the real world. ( For example, the lowest temperatures still have to happen at night and the highest during the day)
1) The minimum and maximum must remain the same, so the amplitude of the graph T(t) = 36.8-1.3sin((Pi/12)(t+2)) should remain 1.3, and 36.8 also neither increases nor decreases. 2) We also should not perform any horizontal shifts of the function. Otherwise, it is possible that we might end up with the highest temperatures at night, and the lowest over the day. So, the phase shift of (Pi/12 * 2)=pi/6 must also remain the same. 3) Since the Earth is now rotating 3 times faster as it used to, we conclude that the only thing that changes is the frequency of equation, which is now 3 times higher. So (Pi/12) should be multiplied by 3. Thus, it becomes (Pi/3) Hence, our final equation is: T(t) = 36.8-1.3sin((Pi/3)t+(Pi/6))
Suppose balanced chemical equation of lightening a match is: 16KClO3 + 3P4 S3 --> 16 KCl + 9SO2+6P2O5 What color must the fire have if all the energy is, somehow, released as light only? State if the answer makes sense or not. Enthalpies of the compounds formations: 1) KClO3 = -392.1 Kj/mol 2) P4S3 = -154.4 Kj/mol 3)KCl = - 411.12 Kj/mol 4) SO2 = -296.8 Kj/mol 5) P2O5 = -1470. Kj/mol
1) Calculate the difference between standard enthalpies of formation between the reactions and products (don't forget to take coefficients into account) You'll get -11330 kJ/mol = -11330000 J/mol 2) Since all of the energy was magically released as light, we can apply the Plank equation to find the released photon wavelength: E=hv=hc/(lambda), where E = energy in Joules, h - Plank constant (6.626 * 10^-34), v - wavelength frequency in Hertz, c = speed of light (3*10^8 m/sec), lambda = wavelength in meters lambda= hc/E = ((6.626*10^-34)(3*10^8))/11330000 = 1.75*10^-32 meters 3) Look at the light color vs wavelength color in your textbook, and find that this wavelength is way outside the visible light range. In addition, even the most energetic gamma-rays have wavelengths of around 10^-12 meters, which is much higher than 1.75*10^-32 meters Thus, the answer does not make sense. And this is consistent with our knowledge of general chemistry. The differences in standard enthalpies of formation results in heat change, not photon absorptions/emissions. The light is released or being absorbed as a result of electrons changing quantum orbitals. When we strike a match, the high temperatures, resulting from friction between the match and the matchbox, and the chemical reaction of combustion, allow electrons to jump back and forth between high and low energies states, which results in the observable light. But its wavelengths are in fact around 6*10^-7 meters, not 1.75*10^-32. We know it because we can see orange-yellow color.
How can one find out an approximate number of leaves in a given tree with: 1) A photo camera 2) Average leaves density inside the tree (d leaves/m^3) 3) Height of the tree (h meters) 4) Radius of the base of the tree
1) Take a picture of the tree and print it 2) Put the picture of a tree into a 2D coordinate system. Let the base of the tree be on the x-axis and its trunk on the y-axis 3) Assume the tree is symmetrical around y-axis 4) Apply linear, quadratic, circular or any other model to the border between the surface of the tree and the rest of the picture, and get equation of the borderline (from x=0 to x where the border crosses the x-axis) using the height of the tree (h) and the radius of the base (r) 5) Calculate definite integral of the obtained function in (4) multiplied by 2(pi)x (cylindrical shell method to calculate the volume of the object). 6) multiply the obtained volume in (5) by density d. This is the number of leaves in the tree