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Garfield

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About Garfield

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  1. Quite easy to prove with similar triangles.
  2. Let's say we have [math]m[/math] kg of oxygen at [math]T_1[/math] K. After heating the oxygen, it did [math]A[/math] J of work. Find the temperature to which the oxygen was heated. [math]p=const[/math]. This is all simple of course, using the [math]A=p(V_2-V_1)=\frac{m}{M}R(T_2-T_1)[/math] formula. Now we are able to find how much did [math]U[/math] change using the [math]U=\frac{3m}{2M}R(T_2-T_1)[/math] formula. Since [math]Q=U+A[/math] according to the first law of thermodynamics, [math]Q=\frac{5m}{2M}R(T_2-T_1)[/math]. Here's the problem: [math]Q=cm(T_2-T_1)=\frac{5m}{2M}R(T_2-T_1)[/math]
  3. 13- similar triangles, find k. 14- stick triangel CDB into ACD, similar triangles, find k.
  4. Ahem...[math] S_{triangle}=\frac{3*4}{2}=6 [/math] Don't you just hate it when that happens when doing calculations in your head.
  5. I came up with a cheap formula to solve the surfice of any triangle. Though I'm not sure if it is correct. I wonder if you 1337 guys could check it.*points to sig*
  6. The thing you'll be doing in the next few minutes: Find a 9-digit number(natural) that has exactely one 3, one 7, one 8 and no 0's in it and when 3 is deleted the number is dividable by 78. When 7 is deleted the number is dividable by 38. When 8 is deleted the number is dividable by 73.
  7. I don't know if I understand this exrecise correctly(bad linguistic skills) but maybe this can help: [math]Q=\rho*m(T_1-T_2)[/math] and [math]N=\frac{A}{t}[/math]. So, if there's 9800kg of substance flowing through in 1h, you need [math]Q=\rho*9800(15-0) J[/math] of energy(note the K's and C's, though) and you need [math]N=\frac{\rho*9800(15-0) }{3600s}W[/math] of power, which probably is constant. [math]\rho[/math] is the Specific Heat Capacity([math]1040\frac {J}{kg*K}[/math] in this case. I'm not sure if this is what you want.
  8. I'm having doubts about 12.5 ft/s... Post height = [math]H[/math] Object height = [math]h[/math] Shadow speed = [math]V[/math] Object speed = [math]v[/math] [math]\frac{H}{h}=\frac{V}{V-v} =>V=\frac{Hv}{H-h}[/math] Then it should be [math]8.(3)\frac{ft}{s}[/math] Is this correct?
  9. [math]\frac{15*5}{6} = 12.5\frac{ft}{s}[/math]?
  10. [math]F_{net} = 0N[/math] because [math]a = 0ms^{-2}[/math], the forces on the beam are [math]F_{grav}=-(F_{rope1}+F_{rope2})=9000N[/math], and the free-body diagram didn't help me either. I'm awful in physics but if someone could give me more hints...
  11. This is a simple exercise but I can't solve it... A 10m 9000N beam is risen up by two parallel wire ropes (a=0m/s^2), one attached to one end and the other 1m from the other end. At what force each wire rope is rising the beam? Sorry for the broken English.
  12. No 5s orbital? And how does the g orbital look like? I know, I'm a new-be. We're only learning "basics" at school.
  13. Orbitals get "filled" with electrons starting from the orbital that has the lowest energy- 1s, than 2s<2p<3s<3p<4s<3d<4p<4d(?)... How does the "sequnce" qontinue?
  14. Sorry if I put this in the wrong sub-forum (I don't know all the mathematical expressions because I study maths in Estonian). Anyways... Two dices are tossed at the same time. What are the chances that the sum of the points is atleast 4? 31/36?
  15. Hello, I'm Garfield. I'm 16 years old and I like science.
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