Recent content by DevonZA
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Hydraulic jack - Pascal's principle
A1= pi/4 * d2 = pi/4 * (0.15)2 = 0.0177m2 A2 = pi/4 * d2 = pi/4 * (0.05)2 = 0.00196m2 Ratio = 0.0177/0.00196 = 9 Therefore ratio is 9:1- DevonZA
- Post #7
- Forum: Engineering and Comp Sci Homework Help
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Hydraulic jack - Pascal's principle
3:1- DevonZA
- Post #5
- Forum: Engineering and Comp Sci Homework Help
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Hydraulic jack - Pascal's principle
I only want to confirm that my answers are correct. In 1.2 I believe they want to know the distance moved on the side of the mass when 10mm pushed down on the small piston side- DevonZA
- Post #3
- Forum: Engineering and Comp Sci Homework Help
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Hydraulic jack - Pascal's principle
Homework Statement Homework Equations P1=P2 P=F/A F=PA F1/A1=F2/A2 F2 = F1(A2/A1) F2 = W(A2/A1) W=mg W=F*d The Attempt at a Solution A1= pi/4 * d2 = pi/4 * (0.15)2 = 0.0177m2 A2 = pi/4 * d2 = pi/4 * (0.05)2 = 0.00196m2 P1=P2 P=F/A F=PA F1/A1=F2/A2 F2 = F1(A2/A1) F2 = W(A2/A1)...- DevonZA
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- Hydraulic Pascal's principle Principle
- Replies: 6
- Forum: Engineering and Comp Sci Homework Help
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Centre of gravity, moment of inertia and radius of gyration
The formulae given for a rectangular triangle: about aa axis I=1/18md^2 about bb axis (through G) I=1/18mh^2 because we want the inertia about xx axis I used =1/18md^2 where d=distance to xx axis=200mm- DevonZA
- Post #9
- Forum: Introductory Physics Homework Help
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Centre of gravity, moment of inertia and radius of gyration
Okay so then X=m1x1-m2x2/m1-m2 = 8*(500)-1.2*(600)/8-1.2 = 482.35mm Is this correct? What about my answers for 1.2 and 1.3?- DevonZA
- Post #7
- Forum: Introductory Physics Homework Help
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Centre of gravity, moment of inertia and radius of gyration
Meaning they should be positive? If they were below the X axis then they would be negative? In other words if the XX axis were at the top of the rectangle.- DevonZA
- Post #5
- Forum: Introductory Physics Homework Help
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Centre of gravity, moment of inertia and radius of gyration
That should be 500, from the centre of the rectangle to the XX axis is 500mm. I believe the centre of the triangle should be 2/3*600 = 400mm Then added to the 200mm from the start of the triangle to the XX axis should give 400+200=600mm. I put a negative in front of the 400 and 1000/3 as they...- DevonZA
- Post #3
- Forum: Introductory Physics Homework Help
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Centre of gravity, moment of inertia and radius of gyration
Homework Statement Homework Equations Centre of gravity: X=m1x1-m2x2/m1-m2 MOI rectangle: 1/3ml^2 MOI triangle: 1/18md^2 Radius of gyration: Ixx=mk^2 The Attempt at a Solution Mass of body 1: b*l*p = 0.8*1*10=8kg Mass of body 2: 1/2b*h*p = 1/2(0.4)*0.6*10=1.2kg1.1 X=m1x1-m2x2/m1-m2...- DevonZA
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- Centre of gravity Gravity Inertia Moment Moment of inertia Radius Radius of gyration
- Replies: 10
- Forum: Introductory Physics Homework Help
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How Do You Calculate the Inside Diameter and Mass Savings for Hollow Shafts?
Homework Statement A solid and hollow shaft of the same material must transmit the same maximum torque. The diameter of the solid shaft is 203mm and the outside diameter of the hollow shaft is to be 216mm. a) Determine the inside diameter of the hollow shaft. (answer = 138.7mm) b) What...- DevonZA
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- Circular Torsion
- Replies: 1
- Forum: Introductory Physics Homework Help
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Change of internal energy of an ideal gas
Thank you let me give that a try.- DevonZA
- Post #5
- Forum: Introductory Physics Homework Help
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Change of internal energy of an ideal gas
Is this because (a-Ru) is a constant? eg. ∫1 dx = x- DevonZA
- Post #3
- Forum: Introductory Physics Homework Help
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Change of internal energy of an ideal gas
Homework Statement Homework Equations Δu = ∫ [(a-Ru)+bT+cT^2+dT^3]dT The Attempt at a Solution The answer of 6447kJ/kmol is given but I am struggling to get to this answer after integrating the above formula and inserting the given values. Firstly would the integral of...- DevonZA
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- Change Energy Gas Ideal gas Internal Internal energy
- Replies: 5
- Forum: Introductory Physics Homework Help