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  1. J

    Thermo Problem, desperate

    Ok this is what I got (I really appreciate the help by the way): CO2=44 g/mole H2O=18 g/mole O2=32 g/mole N2=28 g/mole Total mass of exhaust=565/12 massfractionCO2=.077 massfractionH2O=.064 massfractionO2=.113 massfractioN2=.745 Mass flow of exhaust=mass flow air + mass flow fuel =...
  2. J

    Thermo Problem, desperate

    So if I have 25 kg/s of air and air is 21% O2 I have (.21*25kg/s) gives kg/s of O2. Then (.21*25 kg/s)/32 = N_O2? Similarly ((.79*25kg/s)(3.76N2))/28 gives N_N2 How would it work for CO2 for example? Do I find the amount of moles of C, then use the fact that 25% of the moles of oxygen are...
  3. J

    Thermo Problem, desperate

    I have this reaction: CH4+4(O2+3.76) --> CO2+2H2O+2O2+15.04N2 Through other calculations and given information I have found that the mass flow rate of the air (the O2+3.76N2) is 25kg/s and the mass flow rate of the fuel is .87 kg/s. I am having the hardest time finding, using these mass...
  4. J

    Acetylene Combustion equation

    Well my question wasn't with balancing it, I just wanted to confirm what the product were, not how much of each. So CO2, H20 and N2 are the products?
  5. J

    Acetylene Combustion equation

    Acetylene (C2H2) is burned with a stoichiometric amount of air during a combustion procss. Assuming complete combustion, determine the air-fiel ratio on a mass and mole basis. I'm not asking for help on the actual question I just don't know what the basic chemical equation is? C2H2 +...
  6. J

    Very basic current direction question

    So it only flows through one pair of the capacitors, not both?
  7. J

    Very basic current direction question

    Yes they are supposed to be connected...so once you combine the parallel sets on either side and reduce it to two capacitors are those capacitors in parallel or series
  8. J

    Very basic current direction question

    I attached a picture..Im just curious, if your calculating the capacitance on each plate why is the answer .5 if the V=1000 and each capacitor plate is 500 microfarads.. If C=500 for each plate then first you have two sets of parallel plates so it simples to a series with two plates of 1000...
  9. J

    Mass Flow Rate Through 40 cm Tube at 18 m/s

    If I know air is flowing through a tube with diameter of 40 cm at 18 m/s, with P=1 atm, T=32 C how do I find the mass flow rate..thank you
  10. J

    Mixture Component Problem

    1. A mixture at T=300 K and P=100 kPa is made up of .1 oxygen, .2 carbon dioxide, and .7 Nitrogen. It is then heated to 300 kPa and 500 K. Find the mass of each component I originally thought using PV=NRT, ideal gas but there doesn't seem to be enough givens. Can you also find the molar...
  11. J

    Acceleration on an inclined plane

    This is more a general case question than a specific problem. Say you have a box on an inclined plane and you break the forces into the components, sum the forces in Y and X. If you keep your X and Y axis' the way they usually are (don't make the X axis along the inclined plane) and the box is...
  12. J

    Ideal gas law problem

    Yes but I am still confused. How do you use the mole ratio to find the pressure after heating it up
  13. J

    Ideal gas law problem

    V1=V2=23.2 m^3 or you mean the volume of the tank is constant because I thought the 23.2 was thevolume of the gas not the tank
  14. J

    Ideal gas law problem

    1. A rigid tank contains .5 kmol of Ar and 2kmol of N2 at 250 kPa and 280 K. The mixture is now heated to 400 K. Determine the volume of the tank and the final pressure of the mixture. 2. PV=NRT Can you find the initial volume by using PV=NRT? V=NRT/P V=(2.5)(8.314)(280)/250=23.2...
  15. J

    Thermo Question - Fridge cycles

    Consider a 300 kJ/min refrigeration system that operates on an ideal vapor-compression cycle with refrigerant-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 140 kPa and is compressed to 800 kPa. Determine the (a) quality of the refrigerant at the end of...
  16. J

    Line/Surface Integrals

    I just want to verify how/what each of these is used for: Fundamental Theorem for Line Integrals - This is like the regular fundamental theorem but you use the gradient of F? And this is used for curved lines Greens Theorem - This is only used for simple enclosed curves Stokes' Theorem...
  17. J

    Heat Gain in a Fridge Experiment

    The following is an experiment I have to do: Using a thermometer, measure the temperature of the main food compartment of your refrigerator, and check if it is between 1 and 4°C. Also, measure the temperature of the freezer compartment, and check if it is at the recommended value of -18°C...
  18. J

    Average Rate of Heat Gain in Fridge

    Just curious if anyone knows an approximate for the average rate of heat gain for a common fridge? I am going to do an experiment to calculate the average rate of heat gain in my fridge and want to know if my numbers in the ballpark
  19. J

    Double/triple intergral bounds

    Thats the part I don't understand. How do you know which is which, which is upper and which is lower.
  20. J

    Double/triple intergral bounds

    Hello, I am having trouble understanding how to determine what is the lower and what is the upper bound in some calculus problems. For example: Evaluate the double integral xydA, where D is the region bounded by the line y=x-1 and the parabola y^2=2x+6. Now you set it up to take the...
  21. J

    Merry Go Round

    oh wow whoops so w=.917 so converting back we take (.17 * 60)/2*pi giving 8.75 rev/min. Thanks for the help
  22. J

    Merry Go Round

    Right sorry I always forget..it says I am off by a power of 10, why is this
  23. J

    Merry Go Round

    whoops, r should be squared right...so (23)(1.2^2) = 33.12 So 230(1.05)=263.12w w=,874?
  24. J

    Merry Go Round

    So the moment of inertia for a particle is Mi*ri = (23)(1.2)=27.6 So I of the system is (230+27.6)=257.6. So now we take conservation of angular momentum: L initial = L Final L Initial = I*angular speed = 230*1.05 (10 rev/min = 20pi/min = 1.05rad/s?) L Final = 257.6*angular speed, so we...
  25. J

    Merry Go Round

    A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 230 kgm2 and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the...
  26. J

    Moment of Inertia Door

    Its not supposed to use calculus...I wasn't sure which moment of inertia would apply..i was thinking maybe 1/3MR^2 but not sure
  27. J

    Moment of Inertia Door

    Are you sure, the correct answer said the height wasnt needed?
  28. J

    Moment of Inertia Door

    A uniform, thin, solid door has a height of 2.2 m, a width of 0.87 m, and a mass of 23 kg. Find its moment of inertia for rotation on its hinges. Are any of the data unnecessary? the width of the door is unnecessary the mass of the door is unnecessary no; all of the data is necessary the...
  29. J

    Finding the Moment of Inertia of a System

    so (4)(3^2)+(2)(-2^2)+(3)(-4^2) = (4)(9)+(2)(4)+(3)(16)=36+8+48=92?
  30. J

    Rotating Wheel Problem

    A grinding wheel is in the form of a uniform solid disk of radius 7.01 cm and mass 1.90 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.591 Nm that the motor exerts on the wheel. (a) How long does the wheel take to reach its final rotational...
  31. J

    Finding the Moment of Inertia of a System

    Rigid rods of negligible mass lying along the y-axis connect three particles (Fig. P10.20). The system rotates about the x-axis with an angular speed of 1.20 rad/s. (a) Find the moment of inertia about the x axis. So isn't the moment of inertia given by the sum of mr so...
  32. J

    Chain Problem

    Yes the force from each chain minus the chain should be zero because it is in equilibrium..so do you find the vertical force from the chain then that combined with the forces from the hooks is zero?
  33. J

    Chain Problem

    A flexible chain weighing 42.0 N hangs between two hooks located at the same height (Fig. P12.19). At each hook, the tangent to the chain makes an angle = 41.5° with the horizontal. (a) Find the magnitude of the force each hook exerts on the chain. (b) Find the tension in the chain at its...
  34. J

    Bracket Problem

    Torque around the bottom would equal 0 since it doesn't rotate so: 0=(86)(.05)-(.06)(Fb)?
  35. J

    Bracket Problem

    Ok that makes sense but is this how some of the equations are set up...The force from the screw should be opposing the 86N force, right? Is the bottom of the bracket also opposing the 86N force?
  36. J

    Bracket Problem

    Thats all that is given, the diagram and that question. If its assumed it doesn't rotate the force being applied is pushing it counterclockwise so the bracket must hold it thus pushing clockwise?
  37. J

    Car Center of Mass Problem

    Ahhhh t=(1.5)(14,406)-(2.7)(Ff) Got it now, thanks
  38. J

    Car Center of Mass Problem

    Yes but what is the force..for example if you use the back wheels as the point of rotation the torque is zero as you said. Isn't the torque equation then: 0=(2.7)(Ff) The 2.7 is the distance to the front wheels, the Ff is the force on the front wheels. The torque from the back wheels is...
  39. J

    Bracket Problem

    A shelf bracket is mounted on a vertical wall by a single screw, as shown in Figure P12.59. Neglecting the weight of the bracket, find the horizontal component of the force that the screw exerts on the bracket when an F = 86.0 N vertical force is applied as shown. Ok so we have three...
  40. J

    Car Center of Mass Problem

    A 1470 kg automobile has a wheel base (the distance between the axles) of 2.70 m. The center of mass of the automobile is on the center line at a point 1.20 m behind the front axle. Find the force exerted by the ground on each wheel. A bit confused here. Were going to use the sum of the...
  41. J

    Theory Question

    Well I am trying to derive the equation for acceleration to help explain why it increases and I am having problems doing that. For the hanger: Sum of Forces X = 0 Sum of Forces Y = T-mg=ma Car: Sum of forces X= T=ma Sum of forces Y= 0 I know that much but I am having issues deriving...
  42. J

    Theory Question

    Ok I have a question then when solving for the forces. First the forces of the hanger that's falling The sum of the forces = T-mg=ma assuming down is negative, up is positive. Sum of forces on the car being dragged in the Y direction is 0. Assuming this is frictionless is the sum of the...
  43. J

    Theory Question

    I have a question dealing with acceleration. A car is on a frictionless ramp. The car is tied to a string that goes over a pulley and over the pulley there is a hanger. As the mass of the hanger increases should the acceleration of the car increase, decrease or stay the same. I thought it...
  44. J

    Problem with these types of integrals

    For some reason I just have a problem with these types of integrals..if someone could show me how to do one it would save me a lot of headaches.. 1) x+2/(x^2-4) from 0 to 1 This just simplifies to 1/(x-2) so this should be ln(1-2)-ln(0-2)=ln(-1)-ln(-2)? 2) x/x+1, x/x^2+2...Do you use by...
  45. J

    Rocket Propulsion?

    Yes its webassign..It wants it in metric tons..ive used 3/5 and 2/5 on the first and second respectively
  46. J

    Rocket Propulsion?

    Thats what I originally thought and put it as an answer but it says its wrong
  47. J

    Incline Issue

    Ok so the PE for the 104 block is (104)(9.8)(-20) So: 0=PE+KE+Work 0=(104)(9.8)(-20)+(48)(9.8)(cos37)(20)+(.25*cos(37)*9.8*48*20)+KE This gives: KE=10,992 J Now to find it for the 48kg block we do: (.5*48*v^2)+(.5*104*v^2) = 10,992 Solve for velocity then just plug it into...
  48. J

    Rocket Propulsion?

    Obviously -31,000...but its not -31,000 metric tons... Is it the momentum caused from the engine that gives a speed of 2800 m/s...so P=mv -31,000=2800m, m=11.07
  49. J

    Incline Issue

    So initial energy = final energy + Work by Friction The initial energy is 0 The final PE for the 48kg block is then mgh=(48)(9.8)(sin37*20) The final PE for the 104kg block is (104)(9.8)(sin37*20) because it goes the same vertical distance as the other block? The work of friction is...
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