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Mr-Bonsu-an-engineer-walked-round-a-cylindrical-container-4m-high-once-keeping-a-constant-distance-of-1m-from-the-container-If-he-walked-with-a-speed-of-3kmh-1-for-three-minutes-calculate-to-t




Question Number 152101 by otchereabdullai@gmail.com last updated on 25/Aug/21
Mr Bonsu an engineer walked round  a cylindrical container 4m high once  keeping a constant distance of 1m from  the container. If he walked with a   speed of 3kmh^(−1)  for three minutes,   calculate to the nearest whole number  the:  (i) radius of the circle  (ii) volume of the container
$$\mathrm{Mr}\:\mathrm{Bonsu}\:\mathrm{an}\:\mathrm{engineer}\:\mathrm{walked}\:\mathrm{round} \\ $$$$\mathrm{a}\:\mathrm{cylindrical}\:\mathrm{container}\:\mathrm{4m}\:\mathrm{high}\:\mathrm{once} \\ $$$$\mathrm{keeping}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{1m}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{container}.\:\mathrm{If}\:\mathrm{he}\:\mathrm{walked}\:\mathrm{with}\:\mathrm{a}\: \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{3kmh}^{−\mathrm{1}} \:\mathrm{for}\:\mathrm{three}\:\mathrm{minutes},\: \\ $$$$\mathrm{calculate}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{whole}\:\mathrm{number} \\ $$$$\mathrm{the}: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{container} \\ $$
Answered by Olaf_Thorendsen last updated on 25/Aug/21
2π(r+1)^2  = 3 km.h^(−1) ×3 min  2π(r+1)^2  = ((3000)/(60))×3 = 150  r = (√((75)/π))−1 ≈ 4m  V = πr^2 h = π((√((75)/π))−1)^2 ×4 ≈ 190 m^3
$$\mathrm{2}\pi\left({r}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{3}\:\mathrm{km}.\mathrm{h}^{−\mathrm{1}} ×\mathrm{3}\:\mathrm{min} \\ $$$$\mathrm{2}\pi\left({r}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{3000}}{\mathrm{60}}×\mathrm{3}\:=\:\mathrm{150} \\ $$$${r}\:=\:\sqrt{\frac{\mathrm{75}}{\pi}}−\mathrm{1}\:\approx\:\mathrm{4}{m} \\ $$$$\mathrm{V}\:=\:\pi{r}^{\mathrm{2}} {h}\:=\:\pi\left(\sqrt{\frac{\mathrm{75}}{\pi}}−\mathrm{1}\right)^{\mathrm{2}} ×\mathrm{4}\:\approx\:\mathrm{190}\:\mathrm{m}^{\mathrm{3}} \\ $$$$ \\ $$
Commented by otchereabdullai@gmail.com last updated on 25/Aug/21
thank you sir!
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

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