Recent content by moeraeleizhaj

  1. M

    How Do Electrons Impact Screen Pressure in Televisions?

    In a TV, electrons with a speed of 7.31 x 107 m/s strike the screen from behind, causing it to glow. The electrons come to a halt after striking the screen. Each electron has a mass of 9.11 x 10-31 kg, and there are 7.97 x 1016 electrons per second hitting the screen over an area of 2.15 x 10-7...
  2. M

    Ideal gas equation and boyle's law

    The mass of a hot-air balloon and its occupants is 289 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 890 m3. The...
  3. M

    How Much Heat is Needed for a Steel Sphere to Pass Through an Aluminum Plate?

    r2[(1.1)+(12x10^(-6))(1.1)(D(T)1)]=r2[1+(23x10^(-6))(D(T2))] *cancel r2s out [(1.1)+(12x10^(-6))(1.1)(D(T)1)]=[1+(23x10^(-6))(D(T2))] 12x10^(-6))(1.1)(D(T)1)=1-(1.1)+(23x10^(-6))(D(T2)) D(T)1=[-(0.1)+(23x10^(-6))(D(T2))]/[(12x10^(-6))(1.1)] so...
  4. M

    How Much Heat is Needed for a Steel Sphere to Pass Through an Aluminum Plate?

    I meant the original length would be the original diameter
  5. M

    How Much Heat is Needed for a Steel Sphere to Pass Through an Aluminum Plate?

    okay, so the linear expansion formula is D(L)=(alpha)(Lo)(D(T)) Where D(L)=change in lenth alpha=coefficient of thermal expansion Lo=original length D(T)=change in temp In here I can assume that since it's linear expansion then D(L)=change in diameter for both and the original lenths...
  6. M

    How Much Heat is Needed for a Steel Sphere to Pass Through an Aluminum Plate?

    A 1.96-kg steel sphere will not fit through a circular hole in a 0.813-kg aluminum plate, becuase the radius of the sphere is 0.10% larger than the radius of the hole. If both the sphere and the plate are kept at the same temperature, how much heat must be put into the two so that the ball just...
  7. M

    Bernoulli's equation application

    Ohhh, so you mean I had to solve for the depths of the holes relative to the top of the tank and not relative to each other? I had to do it separately so? Btw, thanks very much for the help ^^
  8. M

    Bernoulli's equation application

    o: Ehm. sorry but why are both sides equal to zero?
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    Bernoulli's equation application

    Btw, I'm kind of stuck on the second bit. The finding the ratio part of the question. Thanks ^^
  10. M

    Bernoulli's equation application

    P1+1/2[*rho*(V1)^2]+[rho*g*(y1)]=P2+1/2[*rho*(V2)^2]+[rho*g*(y2)] Right, P1 and P2 are the atmospheric pressures. Since the atmospheric pressure is the same on both holes, the Ps will cancel each other out. Same with the density (*rho*), since the density is the same throughout the whole...
  11. M

    Bernoulli's equation application

    Bernoulli’s Equation In the steady flow of a nonviscous, incompressible fluid of density (*rho*), the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) are related by : P1+1/2[*rho*(V1)^2]+[rho*g*(y1)]=P2+1/2[*rho*(V2)^2]+[rho*g*(y2)] ^That's Bernoulli's...
  12. M

    Calculating Weight to Make Balloon Rise 116m in 17s Using Archimedes Principle

    that the forces are in equlibruim .'. the weight of the balloon=the bouyant force?
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    Bernoulli's equation application

    [Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Hole 1 is the larger hole, and hole 2 is the smaller hole. The center of one of these holes is located 6.61 times as far beneath the surface of the water as the other...
  14. M

    Calculating Weight to Make Balloon Rise 116m in 17s Using Archimedes Principle

    A lighter-than-air spherical balloon and its load of passengers and ballast are floating stationary above the earth. Ballast is weight (of negligible volume) that can be dropped overboard to make the balloon rise. The radius of this balloon is 6.59 m. Assuming a constant value of 1.29 kg/m3 for...
  15. M

    Liquid pressure and circular motion

    A tube is sealed at both ends and contains a 0.0212-m-long portion of liquid. The length of the tube is large compared to 0.0212 m. There is no air in the tube, and the vapor in the space above the liquid may be ignored. The tube is whirled around in a horizontal circle at a constant angular...
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