Tension Leg Platforms

“sum” of the triangle’s sides: b4.0gt = which leads to a quadratic equation for t 2002 + 82 + 11t . b g 2 46724 + 180.4t − 14.8t 2 = 0. We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by θ = tan −1 . FG 82 + 11t IJ = tan FG 151IJ = 37° . H 200K H 200 K −1 83. Using displacement = velocity × time (for each constant-velocity part of the

“sum” of the triangle’s sides: b4.0gt = which leads to a quadratic equation for t 2002 + 82 + 11t . b g 2 46724 + 180.4t − 14.8t 2 = 0. We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by θ = tan −1 . FG 82 + 11t IJ = tan FG 151IJ = 37° . H 200K H 200 K −1 83. Using displacement = velocity × time (for each constant-velocity part of the

“sum” of the triangle’s sides: b4.0gt = which leads to a quadratic equation for t 2002 + 82 + 11t . b g 2 46724 + 180.4t − 14.8t 2 = 0. We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by θ = tan −1 . FG 82 + 11t IJ = tan FG 151IJ = 37° . H 200K H 200 K −1 83. Using displacement = velocity × time (for each constant-velocity part of the

“sum” of the triangle’s sides: b4.0gt = which leads to a quadratic equation for t 2002 + 82 + 11t . b g 2 46724 + 180.4t − 14.8t 2 = 0. We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by θ = tan −1 . FG 82 + 11t IJ = tan FG 151IJ = 37° . H 200K H 200 K −1 83. Using displacement = velocity × time (for each constant-velocity part of the

straight line but not at constant speed. Let the direction of motion be the positive x direction. (a) Express the total displacement of the car for the entire trip: ∆x total = ∆x1 + ∆x2 Motion in One Dimension Find the displacement for each leg of the trip: 53 ∆x1 = vav ,1∆t1 = (80 km/h )(2.5 h ) = 200 km and ∆x2 = vav , 2 ∆t2 = (40 km/h )(1.5 h ) = 60.0 km Add the individual displacements to get the total displacement: ∆xtotal = ∆x1 + ∆x2 = 200 km + 60.0 km = 260…

straight line but not at constant speed. Let the direction of motion be the positive x direction. (a) Express the total displacement of the car for the entire trip: ∆x total = ∆x1 + ∆x2 Motion in One Dimension Find the displacement for each leg of the trip: 53 ∆x1 = vav ,1∆t1 = (80 km/h )(2.5 h ) = 200 km and ∆x2 = vav , 2 ∆t2 = (40 km/h )(1.5 h ) = 60.0 km Add the individual displacements to get the total displacement: (b) As long as the car continues to…

straight line but not at constant speed. Let the direction of motion be the positive x direction. (a) Express the total displacement of the car for the entire trip: ∆x total = ∆x1 + ∆x2 Motion in One Dimension Find the displacement for each leg of the trip: 53 ∆x1 = vav ,1∆t1 = (80 km/h )(2.5 h ) = 200 km and ∆x2 = vav , 2 ∆t2 = (40 km/h )(1.5 h ) = 60.0 km Add the individual displacements to get the total displacement: ∆xtotal = ∆x1 + ∆x2 = 200 km + 60.0 km = 260…

“sum” of the triangle’s sides: b4.0gt = b 2002 + 82 + 11t . g 2 which leads to a quadratic equation for t 46724 + 180.4t − 14.8t 2 = 0. We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by θ = tan −1 . FG 82 + 11t IJ = tan FG 151IJ = 37° . H 200K H 200 K −1 83. Using displacement = velocity × time (for each…

“sum” of the triangle’s sides: b4.0gt = b 2002 + 82 + 11t . g 2 which leads to a quadratic equation for t 46724 + 180.4t − 14.8t 2 = 0. We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by θ = tan −1 . FG 82 + 11t IJ = tan FG 151IJ = 37° . H 200K H 200 K −1 83. Using displacement = velocity × time (for each…

DASHPOT DATA CAPTURE FORM DATA INPUT DOCUMENT DATA INPUT DOCUMENT DATA MODULE DATE DB CONSUMABLE MATERIALS DB DELIVERY INSTRUCTION DB SPECIFICATION FOR DB SPECIFICATION NUMBER DB SPECIFICATION NUMBER DEAD COUPLING DECALCOMANIA DECK (WORKING PLATFORM) DECK SWITCH DEEP FREEZER DEEP REDUCTION TRANSMISSION DEFLECTING PULL.BRACKET DEFLECTOR DEFROSTER NOZZLE DEHYDRATOR DELETED FROM PACKAGE DELETION OF PARTS DELIVERY LINE DELIVERY PIPE DELIVERY PLUNGER DEMAND, CLAIM…