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Solution Manual For Introduction To Engineering Experimentation, 3rd Edition

Solution Manual For Introduction To Engineering Experimentation, 3rd Edition offers step-by-step solutions to help you understand tough concepts with ease.

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1.1CHAPTER 11.1(a)The weight of a person is 175 pounds which is equivalent to 778 Newtons.lbf175*lbfN14482.4= 778 Newtons(b)The magnetic force of 1,200 dynes is equivalent to 0.0027 pounds or 0.012 Newtons.dyn1200*dynN1105−= 0.012 Newtonsdyn1200*dynN1105−*Nlbf4482.41= 0.0027 pounds(c)The force exerted by the spring is 10 Newtons which is equivalent to 1,000,000 dynes or2.248 poundsN10*Ndyn5101−= 1,000,000 dyneN10*Nlbf4482.41= 2.248 pounds1.2(a)The tire pressure is 30 psig which is equivalent to 2.069 bars or 2.041 atm.Note:1psig is considered to be =1psipsig30*psiatm696.141= 2.041 atmpsig30*psikPa1895.6*kPabar1001= 2.069 bars(b)The compressed air at 10 bars is equivalent to 145 psig or 9.87 atm or 295.3 inches ofmercury or 102 meters of waterbars10*barkPa1100*kPapsi895.61= 145 psibars10*barkPa1100*kPapsi895.61*psiatm696.141= 9.87 atmbars10*barkPa1100*kPapsi895.61*psiatm696.141*atmcmHG176*cmin54.21= 295.3 in of Hgbars10*barkPa1100*kPapsi895.61*psiatm696.141*atmcmWater12.1033*cmm1001= 102 meters water(c)50 cm of water is equivalent to 0.0490 bars or 0.711 psi or 1.45 inches of mercurycmWater50*cmWateratm2.10331*atmkPa1325.101*kPabar1001= 0.0490 barscmWater50*cmWateratm2.10331*atmkPa1325.101*kPapsi895.61* = 0.711 psicmWater50*cmWateratm2.10331*atmcmHG176*cmin54.21= 1.45 in of Hg

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Subject
Mechanical Engineering
Textbook
Introduction To Engineering Experim...

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