Recent content by newyork

Homework Statement Which of the following statements about a Carnot engine is true? a) Any Carnot engine has an efficiency of 100% b) Irreversible energies have the greatest possible efficiency. c) Heat can spontaneously flow from a cold reservoir to a hotter reservoir. d) If a process is...

You would take the difference of those values and divide by 2.

-0.0024m is the total range of length change represented. sorry, I did make a typo before. -25 C to 45 C. the 20 C is right. the 24 should be 25.

Homework Statement Concrete highway slabs are poured in lengths of 10.0 m. How wide should the expansion gaps between the slabs be at a temperature of 20 degrees C to ensure that there will be no contact between adjacent slabs over a temperature range of -24 C to 45 C? Homework...

Thank you so much for the help. That equation made it a lot clearer. I'm going to show my teacher the work tomorrow and see if she approves.

I was quite interested in this discussion because I do research with zinc oxide and titanium dioxide nanoparticles, which are found in cosmetics and sunscreen. Besides being harmful to your circulatory system, they are also hazardous to skin cells. I've tested these nanoparticles mainly on...

Thanks for the help on part b. I understand it now. For part a: "mass of polonium-210 nucleus -> mass of lead-2?? nucleus + mass of alpha particle + alpha particle's KE (+ negligible lead nucleus recoil KE)." This convservation equation confuses me because of the different units...

Before you said, "One electron-volt (eV) is equal to 1.60·10^-12 J" but I thought 1 eV=1.6x10^-19 J. So, I did (5.5x10^6 eV)(1.6x10^-19J)= 8.8x10^-13 J Then, I converted the mass of the alpha particle (4.0026 amu) to kg and got 6.62x10^-27 kg. Finally, I plugged in the bold numbers into...

so using the new KE (I mean it converted to ev), I got v=1.66x10^3. Did I use the right mass, or should I have used the mass I got in part a?

Homework Statement A polonium nucleus of atomic number 84 and mass number 210 decays to a nucleus of lead by the emission of an alpha particle of mass 4.0026 atomic mass units and kinetic energy 5.5 MeV. (1 atomic mass unit=931.5 MeV/c^2=1.66x10^-27kg.) The atomic number of the lead nucleus...