1)A 3000-lb wrecking ball hangs from a 30-ft cable of density 9 lb/ft attached to a craneCalculate the work done if the crane lifts the ball from ground level to 30 ft in the air by drawing in the cable.I've tried simply doing 30 x 9 x integral of y dy from 0 to 30 but to no avail2)A circular swimming pool has a diameter of 16 m, the sides are 3 m high, and the depth of the water is 2.5 mHow much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 and the density of water is 1000 .) I've tried many scenarios but the main problem is that its only half of a circleThanks in advance!
Hi: The reason is that the tube insides is coated with a thin film of teflon which the stuff can't stick too, due to the fact that teflon is chemically inert ( or can't bond to anything chemically) if you can read it on the instruction for crazy glue: do not use it to bond teflon or other florialcarbon compoundsThat is why you can't glue those compoundsTeflon or any other teflon-like compounds on any cookware is mechanically bonded on ,not chemically Hope this helps.
Because while it is wet it is not bonded and by keeping the cap on, you keep the air out and prevent it from drying and bondingIf you were the leave the cap off of a tube of super glue, it would dry and stick to the inside of the tube,
You just need to set up the integrals properlyIn the first problem, you're given: - Mass of the ball, Mb - Density of the cable, pc - Length of the cable, lc At any given point, the amount of mass hanging off the edge of the crane is: M(y) (Mb + pc y) where y is the amount of cable that has been drawn inYou know that work is equal to the integral of force applied through a pathThe force due to gravity at any given point is simply: F(y) M(y) g (Mb + pc y) g Thus, the work done is simply the integral of F(y) dy over 0 y lc: W int[(Mb + pc y) g dy] from 0 to lc This gives a final expression of: W Mbglc + 1/2 pc g lc^2 Recognizing that since Mb is given in pounds (which is a force, not a mass), we can drop the g termThis gives: W 3000lb 30ft + (1/2) 9 lb/ft (30 ft)^2 W 94050 ft-lbs or ~ 127.5 kJ - In problem 2, I think you're over-complicating the problem statement (there may not be an accompanying picture)The way I read it, the pool is a cylinder of depth 3m and diameter 16mThat is, the bottom of the pool is level, not curvedI think your assumption that there's a semi-circle involved is inaccurateThe infinitesimal amount of work done in raising an infinitesimal amount of water is: dW dM g h The infinitesimal mass of water raised is density times its infinitesimal volume, which in this case is: dM p dV dM p A dh where A is simply the cross-section of the pool: A pi r^2 A (1/4) pi d^2 Putting it all together, you get: dW (1/4) p pi d^2 g h dh Finally, integrate dW over 0.5m dW 3m (the very top layer of water only needs to be raised 0.5m, while the lowest layer of water needs to be raised the full 3m)Doing so gives: W (35/32) p pi d^2 g When you input all of your known values (p1000kg/m^3, d16m and g9.8m/s^2), you get: W 8,620,530 J Hope that helps
I have always wondered that I assume it has something to do with it's reaction to oxygen that make it stick, since the tube is mostly air tight.
Generally, cyanoacrylate is an acrylic resin which rapidly polymerises in the presence of water (specifically hydroxide ions), forming long, strong chains, joining the bonded surfaces togetherBecause the presence of moisture causes the glue to set, exposure to moisture in the air can cause a tube or bottle of glue to become unusable over timeTo prevent an opened container of glue from setting before use, it must be stored in an airtight jar or bottle with a package of silica gel.
