Water flows steadily from an open tank the elevation of point 1 is 10.0 m , and the elevation of points 2 and 3 is 2.00 m . the cross-sectional area at point 2 is 4.80×10−2 m2 ; at point 3, where the water is discharged, it is 1.60×10−2 m2 . the cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. assuming that bernoulli's equation applies, compute the volume of waterδv that flows across the exit of the pipe in 1.00 s . in other words, find the discharge rate δv/δt. express your answer numerically in cubic meters per second.δvδt =
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Physics, 22.06.2019 20:30, valenzueladomipay09u
Light from two different sources are incident upon the same side of a polarizing filter. light from one source is polarized with total intensity ip, while light from the other source is unpolarized with total intensity iu. as you rotate the polarizing filter through 360 degrees, the intensity of the transmitted light on the other side varies by a factor of 9. what is the relative intensity of the initial polarized source to the unpolarized source, or ip/iu?
Answers: 2
Water flows steadily from an open tank the elevation of point 1 is 10.0 m , and the elevation of poi...
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