Question

A 0.504 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plunger. The cylinder comes to rest some distance above the plunger. The plastic tube has an inner radius of 7.24 mm and is frictionless. Neither the plunger nor the metal cylinder allow any air to flow around them. If the plunger is suddenly pushed upwards, increasing the pressure between the plunger and the metal cylinder by a factor of 2.07 , what is the initial acceleration ? of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm and that the top of the tube is open to the air.

Answer 1

Sól? Given: Mass of the metal cyltoscler,m= 0.504 Kg Inner radius of plastic tube, & = 7624mm = 7,24x10m factor f = 2.07 pressure outside the fube Pa = 1.00 atm = 1.01x100Pa the pressure inside the tube P, = Pat w "Wang & weight of cylinches A=722 Area of plastic tube = Pat mg 772 - 1.01 Xlornat dives . 0.504x9.8 * (7.24x15382 lol xlopat 4.9.39 kg. m/s2 IP4 P, 1.646870-4m2 hete - 1.01 Xlorat 30006.07 kg/m3? x = ..^2 kg/m2 = 131006007 pa pressure applied by the plunger P = fp P2 = 2.07 x 131006.07 P2 = 271182.568 P4 predsybe clifference ops Pe-P, OP = 271182.564131006. 07 DP = 140176.49 alow force applied by the metal cylinder Scanned with FosPxA CamScanner

F = 140176.49 X 10646 X10-9 F = 23.070 e acceleration of the metal cylinder Fama 23.07 - 00500 xq q- 23.07 ooray la = 45.77 mls?) Am if you are satisfy with us then please appreciate us with a like. If not then tell us in comment box we are happy to help you. again thank you Ics Scanned with CamScanner