A hydraulic lift has two connected pistons with cross-sectional areas 20 cm2 and 290 cm2.?
610131
x/20 = 1500/290\n\nx =103.45kg \n\n2) hd = 90/290 where d = .0005 kg/ cm^3\nh =620.69 cm.\n3) 290/ 620.69 = 0.4672 cm
637941
3 feet and 2 1/2 inches
1097483
P/T=P/T\n1/(20+273)=P/(600+273)\nP=30 atm\n\nThe new pressure would be 30 atms, but this is considering that the volume remains constant.
831548
V = 2hw², h = 48/w²\nAb = 2w²\nAs = 2wh\nAe = wh\nAl = 2(2wh + wh) = 6wh = 6*48/w\nC = 12*2w² + 5*6*48/w\ndC/dw = 48w - 1,440/w² = 0\n48w = 1,440/w²\nw³ = 30\nw = 3.1072 m\nh = 48/3.1072² \nh = 4.9716 m\nC = $781.44
533825
Calculate the cross-sectional area of the water and multiply by the length of the cylinder.  The cross sectional area can be calculated with the depth of the wafer and the radius of the cylinder.  It is hard to explain without drawing a diagram but basically you draw 2 straight line connecting the center of the cycle (cross-section) to where the water surface contact the wall of the cylinder.  That form a sector.  Calculate the angle of the sector.  Multiply the total area of the circle by the angle of the sector divided by 360 degree to obtain the area of the sector.  Next, substract the area in the sector where there is no water and you will get the cross-sectional area of the water.
380811
160.02 centimeters\n\n1.60 meters\n\n1 meter and 60 centimeters
114490
u may want to try Handbook of Hydraulic Fluid Technology\nMechanical Engineering (Marcel Dekker, Inc.) by George E. Totten ISBN: 0824760220
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
752166
Let F be the force applied to the pump handle\n\n0.35m x F = 0.04m x 400N\n=> F = 45.7N
396424
The curved surface area of the cylinder would be made up of the circumference times the height.  Figure the circumference then divide the surface area by that to get the height.  it would probably be best to convert M2 into cm2 first.  100 cm=1M so 1 sq m would be 100 x 100 or 10,000sq cm x 4m2 equals 40000 cm2.  The C would be 2(pi)r  which is 2 x 3.14 x 70 = 439.6 cm.  Divide 40,000 by 439.6 = 90.99 cm for the height. volume would be (pi)r2h or 3.14 x 4900 x 90.99 = 1399972 cu cm
762826
Hydraulic machines work on Pascals principle.\nAccording to it the pressure exerted on a confined fluid is tranmitted equally and undiminshed in all directions.\nSo we can apply a small force on a large area and that pressure is transmitted throughout the fluid.\nThis pressure can be used to lift relatively large load(or force) by making it act on a small area of cross-section on the platform of the machine.\nPascals Principle is : P1=P2\nor\nF1/A1=F2/A2
1187798
The pressure difference between the opening(s) and the constricted part of the venturi should be proportional to their cross-sectional areas taken as Pi *r^2 or Pi*(D/2)^2.\n\n* is multiply\nPi is ~ 3.14\nr is radius\nD is diameter\n^ is "to the power"\n\nIf the opening has a diameter of D, and the constricted part has a diameter of d, then the pressure increases at d by the multiple of:\n\npi*(D/2)^2  /  pi*(d/2)^2\n\neverything cancels except (D/d)^2\n\nso, if the diameter of the openings is 2 units (like inches) and the diameter of the narrowest part is 1 unit, the pressure at the narrowest point increases by:\n\n(2/1)^2 = 4  or 4-fold.\n\nYou should have someone check me; this is derived from scratch!
139357
Hydraulic technology relies on the fact that liquid, unlike gases and solids, cannot be compressed. Put some water in a bucket and try to push it down. Right—it can't be done. The water is displaced. Imagine an apparatus made of three components: a reservoir divided into two sections (one small, one large) with a passage between the two sections. In each section is a piston. If you fill the reservoir with fluid and then apply pressure to the small piston (called a master cylinder), the large piston will rise.
761667
cylinder\n1.volume=pir^2h\n2.LSA=2pirh\n3.TSA=2pir(h+r)\n\nsphere\n1.volume=(4/3)pir^3\n2.area=4pir^2\n\nhemisphere\n1.volume=(2/3)pir^3\n2.basearea=pir^2\n3.LSA=2pir^2\n4.TSA=3pir^2\n\ncone\n1.volume=(1/3)pir^2h\n2.LSA=pirl\n3.TSA=pir(l+r)\n\ncube\n1.volume=a^3\n2.surface area=6a^2\n\ncuboid\n1.volume=lbh\n2.surface area=2(lb+lh+bh)
1075425
H\n\nArea of first circle = pi*20^2 = 400*pi\nArea of second circle = pi*10^2 = 100*pi\npi(400-100) = 300*pi = 942
908785
very close to 2 meters.
953550
see this website :\nhttp://www.hydraulicspneumatics.com/200/FPE/Accessories/Article/True/6443/Accessories