Question Number 9209 by tawakalitu last updated on 23/Nov/16 Commented by tawakalitu last updated on 23/Nov/16 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9204 by tawakalitu last updated on 23/Nov/16 Answered by mrW last updated on 23/Nov/16 $$\mathrm{300}×\mathrm{9}.\mathrm{81}×\mathrm{20}/\mathrm{60}=\mathrm{981}\:\mathrm{Nm}/\mathrm{s} \\ $$$$\mathrm{The}\:\mathrm{power}\:\mathrm{rating}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pump}: \\ $$$$\mathrm{981}/\mathrm{0}.\mathrm{80}=\mathrm{1226}\:\mathrm{Watt} \\ $$$$\mathrm{The}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{loss}\:\mathrm{of}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{20\%}. \\ $$…
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Question Number 9161 by tawakalitu last updated on 21/Nov/16 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{rectangular}\:\mathrm{plate}\:\mathrm{of}\:\mathrm{lamina}\: \\ $$$$\mathrm{density}\:\rho\:=\:\mathrm{0}.\mathrm{025}\:\mathrm{whose}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{5kg}\:\mathrm{about} \\ $$$$\mathrm{an}\:\mathrm{axis}\:\mathrm{3cm}\:\mathrm{from}\:\mathrm{one}\:\mathrm{of}\:\mathrm{its}\:\mathrm{sides}\:\mathrm{as}\:\mathrm{shown} \\ $$$$\mathrm{below}.\:\mathrm{using}\:\mathrm{parallel}\:\mathrm{axis}\:\mathrm{theorem}\:,\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{about}\:\mathrm{the}\:\mathrm{xx}\:\mathrm{axis}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{gyration}. \\ $$ Commented by tawakalitu…
Question Number 74688 by TawaTawa last updated on 28/Nov/19 $$\mathrm{y}\:\:=\:\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{tranform}\:\mathrm{this}\:\mathrm{into}\:\mathrm{a}\:\mathrm{real}\:\mathrm{life}\:\mathrm{problem}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{with} \\ $$$$\mathrm{several}\:\mathrm{condition}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74655 by TawaTawa last updated on 28/Nov/19 $$. \\ $$ Commented by TawaTawa last updated on 28/Nov/19 The force F acting along an inclined plane is just sufficient to maintain a body on the plane, the angle of friction M being less than Y, the angle of plane. prove that the least force acting along the plane, sufficient to drag the body up the plane is : F sin( M + Y )/sin( M - Y) Answered by mr W last…
Question Number 9101 by tawakalitu last updated on 18/Nov/16 Commented by mrW last updated on 18/Nov/16 $$\left.{b}\right) \\ $$$$\left.{see}\:{a}\right)\:{below} \\ $$$$\mathrm{15}{s}+\mathrm{4}={mv}\frac{{dv}}{{ds}} \\ $$$$\left(\mathrm{15}{s}+\mathrm{4}\right){ds}={mvdv} \\ $$$$\int\left(\mathrm{15}{s}+\mathrm{4}\right){ds}=\int{mvdv}…
Question Number 74632 by TawaTawa last updated on 27/Nov/19 $$. \\ $$ Commented by TawaTawa last updated on 27/Nov/19 The force F acting along an inclined plane is just sufficient to maintain a body on the plane, the angle of friction M being less than Y, the angle of plane. prove that the least force acting along the plane, sufficient to drag the body up the plane is : F sin( M + Y )/sin( M - Y) Terms of Service Privacy Policy…
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Question Number 140143 by I want to learn more last updated on 04/May/21 Commented by mr W last updated on 04/May/21 $$\left({a}\right) \\ $$$${F}_{{W}} =\mathrm{12}×\mathrm{tan}\:\mathrm{30}°\approx\mathrm{6}.\mathrm{9}\:{N}…