Water (density$$ ρ$$)is$$ flowing through the uniform horizontal tube of $$cross-sectional$$ area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is $$(neglect$$ viscous $$forces)

Question

Water (density$$ ρ$$)is$$ flowing through the uniform horizontal tube of $$cross-sectional$$ area  A with a constant speed  v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is $$(neglect$$ viscous $$forces)

Infinity Learn · Accepted Answer

GivenDensity of water $$=$$ ρ$$Cross-sectional$$ $$area=$$ AConstant $$speed=$$ vAlong  x and  y directions along in momentum of water is P→$$i=mvj^Pf$$→$$=mv32i^-mv2j^$$∆P→$$=Pf$$→$$-Pi$$→$$=mv32i^-mv32j^$$⇒ΔP¯$$net=32mv2+-32mv2=3mvForce$$ on bend is rate of change of momentum ⇒ΔF¯$$net=3dmdt$$.$$v=3$$ρ$$Av2$$

Water (density $ρ$ )is flowing through the uniform horizontal tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)

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Water (density ρ)is flowing through the uniform horizontal tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)

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Water (density $ρ$ )is flowing through the uniform horizontal tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)