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It depends on the length of the pipe.
Yes, if the joint between the 2 width of pipes is made to be tightly closed so that no leakage would occur. The best would be to use a connector with a 1 inch diameter at one end and a diameter of 1.5 inches at the other end. Or maybe, if the 2 pipes are vertical, to put the 1 inch pipe into the 1.5 inches pipe and seal the 2 pipes with a good amount of resistant tape so that the water will go straight from one pipe to the other with no leakage. But with hot water draining from the washer, even with the cold water, the tape solution is probably just a temporary solution, because the tape would break down after some time.
assuming that the pipes are all the same length, then you can work it out as follows: The pipes are cylindrical, and the volume of a cylinder is equal to pi multiplied by the square of it's radius multiplied by the length of the pipe - or: v = πr2h we can compare the two then by saying the volume of our two smaller pipes is: v = πr12h + πr22h r1 = 3 and r2 = 4, therefore v = π32h + π42h v = π9h + π16h v = π25h and then we can say that the volume of our biggest pipe is: v = πr2h r = 5, therefore v = π52h v = π25h therefore, the two small pipes carry the same amount as the large pipe
To start, we will find the cross sectional area of the pipes. Pipe with radius of 2 inches Cross sectional area = PI x radius2 = 12.56 inches2 Pipe with radius of 3 inches Cross sectional area = PI x radius2 = 28.26 inches2 Pipe with radius of 5 inches Cross sectional area = PI x radius2 = 78.5 inches2 As the cross sectional area of the pipe with a radius of 5 inches (78.5 inches2) is greater than the sum of the cross sectional areas of the pipes with radii of 2 inches and 3 inches (40.82 inches2), it can be concluded that the pipe with a radius of 5 inches will be able to carry the most water. (Assuming that all three pipes are the same length).
There is no such unit of measurement as the "pipe".