Question Number 68714 by Learner-123 last updated on 15/Sep/19 Answered by mr W last updated on 15/Sep/19 $${acceleration}\:{of}\:{object}\:{C}=\frac{\mathrm{1}.\mathrm{2}}{\mathrm{2}}=\mathrm{0}.\mathrm{6}{m}/{s}^{\mathrm{2}} \\ $$$${force}\:{in}\:{cable}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{375}×\frac{\mathrm{10}+\mathrm{0}.\mathrm{6}}{\mathrm{10}}=\mathrm{198}.\mathrm{75}\:{N} \\ $$$${power}\:{input}=\frac{\mathrm{198}.\mathrm{75}×\mathrm{0}.\mathrm{6}}{\mathrm{0}.\mathrm{85}}=\mathrm{140}.\mathrm{3}\:{watt} \\ $$ Commented…
Question Number 134249 by abdurehime last updated on 01/Mar/21 Answered by mr W last updated on 01/Mar/21 Commented by mr W last updated on 01/Mar/21…
Question Number 68642 by Maclaurin Stickker last updated on 14/Sep/19 $$\mathrm{Young}'\mathrm{s}\:\mathrm{modulus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{material}\:\mathrm{measures} \\ $$$$\mathrm{its}\:\mathrm{resistance}\:\mathrm{caused}\:\mathrm{by}\:\mathrm{external}\:\mathrm{stresses}. \\ $$$$\mathrm{On}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{wall}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{specific} \\ $$$$\mathrm{mass}\:\rho\:\mathrm{and}\:\mathrm{Young}\:\varepsilon\:\mathrm{modulus}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{parallelepiped}\:\mathrm{shape},\:\mathrm{the}\:\mathrm{dimensions} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{which}\:\mathrm{are}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\: \\ $$$$\mathrm{Based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{correlations}\:\mathrm{between}\:\mathrm{physical} \\ $$$$\mathrm{quantities},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{that}…
Question Number 134064 by mr W last updated on 27/Feb/21 Commented by mr W last updated on 27/Feb/21 $${find}\:{the}\:{angle}\:\theta\:{at}\:{which}\:{the}\:{falling} \\ $$$${rod}\:{begins}\:{to}\:{slip}\:{on}\:{the}\:{ground}\:{if} \\ $$$${the}\:{friction}\:{coefficient}\:{between}\:{rod} \\ $$$${and}\:{ground}\:{is}\:\mu.…
Question Number 68510 by oyemi kemewari last updated on 12/Sep/19 Answered by mr W last updated on 14/Sep/19 $${v}=\sqrt{\mathrm{2}{gh}} \\ $$$${base}\:{area}\:{of}\:{tanks}={A}=\mathrm{1}\:{ft}^{\mathrm{2}} \\ $$$${A}\frac{{dh}}{\mathrm{2}}=−\frac{\pi{d}^{\mathrm{2}} }{\mathrm{4}}{vdt} \\…
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Question Number 68487 by mr W last updated on 11/Sep/19 Commented by mr W last updated on 11/Sep/19 $${rod}\:{AB}\:{has}\:{length}\:{b}\:{and}\:{mass}\:{M}, \\ $$$${rope}\:{BC}\:{has}\:{length}\:{l}\:{and}\:{mass}\:{m}, \\ $$$${the}\:{friction}\:{between}\:{rod}\:{and}\:{ground} \\ $$$${is}\:{sufficient},\:{such}\:{that}\:{point}\:{A}\:{can}…
Question Number 68451 by mr W last updated on 10/Sep/19 Commented by mr W last updated on 11/Sep/19 $${if}\:{M}=\mathrm{0},\:{the}\:{block}\:{m}\:{just}\:{has}\:{free} \\ $$$${fall},\:{therefore}\:{T}=\sqrt{\mathrm{2}{gh}}=\sqrt{\mathrm{2}{gR}}. \\ $$ Commented by…
Question Number 68418 by necxxx last updated on 10/Sep/19 Commented by necxxx last updated on 10/Sep/19 $${The}\:{UDL}\:{is}\:\mathrm{2}.\mathrm{5}{KN}/{m}.\:{The}\:{first}\:{point} \\ $$$${load}\:{is}\:\mathrm{35}{KN}.\:{I}'{ve}\:{tried}\:{solving}\:{but}\:{I} \\ $$$${encountered}\:{some}\:{issues}\:{with}\:{the}\:{diagram}. \\ $$$${I}'{ll}\:{be}\:{really}\:{grateful}\:{to}\:{get}\:{this}\:{solved} \\ $$$${with}\:{the}\:{possible}\:{diagrams}.…
Question Number 68342 by Tony Lin last updated on 09/Sep/19 Commented by Tony Lin last updated on 09/Sep/19 $${when}\:{m}\:{slide}\:{to}\:{the}\:{bottom} \\ $$$${if}\:{there}\:{is}\:{no}\:{fraction} \\ $$$${find}\:{v}_{{m}} \&{v}_{{M}} \\…