POINT (2, 2, 1) : X = 2; y =2; z =1U̅ = {4 × (2)2}i – {5 × (2)2× 2} j + 1kU̅ = 16i- 40 j+ kThe magnitude of VELOCITY is given by:\(u = \sqrt {{{\left( {16} \RIGHT)}^2} + \;{{\left( { - 40} \right)}^2} + \;{{\left( 1 \right)}^2}} = \;\sqrt {1857\;} \) m/s∴ u = \(\sqrt {1857\;} m/s\)

"> POINT (2, 2, 1) : X = 2; y =2; z =1U̅ = {4 × (2)2}i – {5 × (2)2× 2} j + 1kU̅ = 16i- 40 j+ kThe magnitude of VELOCITY is given by:\(u = \sqrt {{{\left( {16} \RIGHT)}^2} + \;{{\left( { - 40} \right)}^2} + \;{{\left( 1 \right)}^2}} = \;\sqrt {1857\;} \) m/s∴ u = \(\sqrt {1857\;} m/s\)

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A fluid flow is described by velocity field U̅ = 4x2i – 5x2yj + 1kWhat is the absolute velocity (in magnitude) at the point (2, 2, 1)?

BITSAT Kinematics in BITSAT 9 months ago

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U̅ = 4x2i – 5x2yj + 1kAt POINT (2, 2, 1) : X = 2; y =2; z =1U̅ = {4 × (2)2}i – {5 × (2)2× 2} j + 1kU̅ = 16i- 40 j+ kThe magnitude of VELOCITY is given by:\(u = \sqrt {{{\left( {16} \RIGHT)}^2} + \;{{\left( { - 40} \right)}^2} + \;{{\left( 1 \right)}^2}} = \;\sqrt {1857\;} \) m/s∴ u = \(\sqrt {1857\;} m/s\)

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Posted on 16 Nov 2024, this text provides information on BITSAT related to Kinematics in BITSAT. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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