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Garfield

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Posts posted by Garfield

  1. 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]

    from witch

    [math]c=\frac{5R}{2M}[/math]. Numbers don't show that the given formula is correct nor is it correct when we take [math]V=const, c=\frac{3R}{2M}[/math]. So, my question is: Why is the above all wrong, and how do you calculat c with ideal gases and how to make a small triangle appear in front of [math]T[/math]?

  2. 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.

  3. 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.

  4. 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?

  5. [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...

  6. 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.

  7. 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?

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