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A water jet whose cross sectional area is a strikes a wall making an angle theta with the normal and rebounds elastically. The velocity of water of density d is v. What is the force exerted on wall |
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Answer» answer : 2dav²cosθ explanation : given, area of cross sectional of water jet = a density of water = d velocity of water = V mass of water, m = density of water × VOLUME of water = d × (x × a) = dax [ here x is length of water FLOWING through the water jet ] a/c to question, A water jet whose cross sectional area is a strike a wall MAKING an angle θ with normal and rebounds elastically as shown in FIGURE. so, change in momentum, ∆P = mvcosθ- (-mvcosθ) = 2mvcosθ putting, m = dax so, ∆P = 2daxvcosθ force exerted on wall = change in momentum/time = 2daxvcosθ/t = 2davcosθ (x/t) = 2davcosθ (v) = 2dav²cosθ |
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