How can I shorten my dogs tie-out cable?
I just wrap one end around a pole or tree as many times as needed to take up the extra-length I take a 30 ft with me on trips --- sometimes there is only room for 15 ft -- so there is alot wrapped around something If you wrap it first & then put the hook on the line, the dog can't unwrap it -- if you hook it first & then wrap, the dog could just go around the pole or tree & unwrap all of it tieing the line, if it has a metal core will cause it to fray, split & the metal will come through the plastic casing... best to not tie a knot in the line
Where to hook grund wire from pool pump?
**************THERE IS NO GROUNDING LUG ON A POOL THAT IS A BONDING LUG MOST ABOVE GROUND POOL HAVE NOTHING THAT NEEDS TO BE BONDED DO NOT GROUND A POOL MOTOR DO NOT DRIVE A GROUND ROD YOU NEED TO BOND THINGS YOU ARE VERY VERY CLOSE TO KILLING SOMEONE PLZ CALL A PRO I have been involved in 8 hour courses teaching master electricians how to safely wire pools you are not going find anything that can explain it all to a non electrician just call a pro that is very well schooled in pools MUCH ABOUT POOLS IN COUNTER INTUITIVE YOU WILL NOT FIGURE IT OUT EDUCATED GUESS DONT WORK FOR EXAMPLE DRIVING A GROUND RODS OR GROUNDING ANYTHING WITH IN 10-20 FEET OF A POOL KILLS A LOT OF PPL EVERY YEAR the guy who said you the white as a ground cuz the go the same place is a fool and understand nothing about electrical safety. If an electrician did that it would be so unsafe it would be almost criminal
How do I extend my wifi signal to another building 400 feet away?
Does it need to be wireless? Assuming the ranch office has electricity then you might be better off using a powerline modem, and then having a WiFi AP in the office. Or even run a power line just to do this. Little known fact: copper conducts electricity better than air ;)People have got WiFi to go a very long way, but as Bill McDonald says, it isn't quite trivial. There are directional antennas (the famous Pringle can !) which can work surprisingly well, but perhaps best to get help.
Geometry problem: PLEASE HELP SOLVE!?
Since it doesn't matter how far apart the poles are, choose a convenient number, say 10 m apart. Then draw the figure with a vertical segment x representing the height of the intersection. Call the side from that segment's bottom to the bottom of the 20 m pole a, and the side from there to the 30 m base is 10-a By similar triangles 20/10 = x/ (10-a) and 30/10 = x/a cross multiply and 10x = 200 - 20a while 10x also = 30a so 30a = 200 - 20a 50a=200 a=4 Then plug that in to get x = 12
How to replace the power wire a Hayward Power-flo Matrix Pool Pump?
As long as the green wire is on the "ground" screw (which is the green one) you can plug the black and white to either port. I worked on these for many years. Look at the diagram on the motor. It will tell you which ports to use. If the motor diagram is worn off, just go to Hayward's website. www.haywardnet.com and you can download owners manuals there. If you are still apprehensive take it to your local Leslie's. Leslies always offers free labor. Just buy the plug and ask them to put it on. They might have to check it in because they get very busy this time of year, but they won't charge you to put it on.
Two simple physics problems?
You just need to set up the integrals properly. In 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 in. You know that work is equal to the integral of force applied through a path. The 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 = Mb*g*lc + 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" term. This 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 16m. That is, the bottom of the pool is level, not curved. I think your assumption that there's a semi-circle involved is inaccurate. The 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 (p=1000kg/m^3, d=16m and g=9.8m/s^2), you get: W = 8,620,530 J Hope that helps