Question Number 10392 by ketto last updated on 06/Feb/17 $${find}\:{the}\:{direction}\:{cosines}\: \\ $$$${and}\:{its}\:{angles}\:{on} \\ $$$$\mathrm{2}{i}\:−\:\mathrm{3}{j}\: \\ $$ Answered by mrW1 last updated on 07/Feb/17 $$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}}…
Question Number 10369 by Tawakalitu ayo mi last updated on 05/Feb/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\mathrm{2}\:+\:\mathrm{5}\:+\:\mathrm{8}\:+\:\mathrm{12}\:…\:\mathrm{n}\: \\ $$ Commented by mrW1 last updated on 06/Feb/17 $${the}\:{n}−{th}\:{term}\:{of}\:{the}\:{series}\:{is}\:{not} \\…
Question Number 10362 by Tawakalitu ayo mi last updated on 05/Feb/17 $$\mathrm{An}\:\mathrm{helium}\:\mathrm{atom}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{6}.\mathrm{64}\:×\:\mathrm{10}^{−\mathrm{27}} \mathrm{kg}\:\mathrm{and}\:\mathrm{a}\:\mathrm{charge} \\ $$$$\mathrm{Q}\:\mathrm{is}\:+\mathrm{2}\:\mathrm{electon}.\:\mathrm{Compare}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{electric}\: \\ $$$$\mathrm{repulsion}\:\mathrm{to}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{gravitational}\:\mathrm{attraction}\:\mathrm{between}\:\mathrm{them}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 10360 by Tawakalitu ayo mi last updated on 05/Feb/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{5}^{\mathrm{x}} \:=\:\mathrm{6}^{\mathrm{x}} \\ $$ Answered by mrW1 last updated on…
Question Number 10355 by Tawakalitu ayo mi last updated on 05/Feb/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{escape}\:\mathrm{velocity}\:\mathrm{from}\:\mathrm{the}\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{planet}\:\mathrm{with}\:\mathrm{two}\:\mathrm{third}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}'\mathrm{s} \\ $$$$\mathrm{gravity}\:\mathrm{but}\:\mathrm{the}\:\mathrm{same}\:\mathrm{radius}. \\ $$ Commented by Tawakalitu ayo mi last updated…
Question Number 75879 by TawaTawa last updated on 19/Dec/19 Answered by mr W last updated on 19/Dec/19 Commented by mr W last updated on 19/Dec/19…
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Question Number 10309 by Tawakalitu ayo mi last updated on 03/Feb/17 $$\mathrm{An}\:\mathrm{helium}\:\mathrm{atom}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{6}.\mathrm{64}×\mathrm{10}^{−\mathrm{27}} \mathrm{kg} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{charge}\:\mathrm{Q}\:\mathrm{is}\:+\mathrm{2}\:\mathrm{electron}.\:\mathrm{Compare}\:\mathrm{the} \\ $$$$\mathrm{magnitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{repultion}\:\mathrm{to}\:\mathrm{that} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{gravitational}\:\mathrm{attraction}\:\mathrm{between}\:\mathrm{them}. \\ $$ Terms of Service Privacy…
Question Number 75847 by behi83417@gmail.com last updated on 18/Dec/19 $$\begin{cases}{\frac{\boldsymbol{\mathrm{tgx}}−\boldsymbol{\mathrm{tgy}}}{\mathrm{1}−\boldsymbol{\mathrm{tgx}}.\boldsymbol{\mathrm{tgy}}}=\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}}\\{\:\:\frac{\boldsymbol{\mathrm{tgx}}+\boldsymbol{\mathrm{tgy}}}{\mathrm{1}+\boldsymbol{\mathrm{tgxtgy}}}=\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}}}\end{cases} \\ $$ Commented by MJS last updated on 19/Dec/19 $$\mathrm{the}\:\mathrm{path}\:\mathrm{is} \\ $$$$\mathrm{let}\:{x}=\mathrm{2arctan}\:{p};\:{y}=\mathrm{2arctan}\:{q} \\ $$$$\Rightarrow \\…
Question Number 10286 by Tawakalitu ayo mi last updated on 02/Feb/17 $$\mathrm{2000g}\:\mathrm{of}\:\mathrm{water}\:\mathrm{at}\:\mathrm{100}°\mathrm{C}\:\mathrm{is}\:\mathrm{poured}\:\mathrm{into}\:\mathrm{a}\: \\ $$$$\mathrm{copper}\:\mathrm{calorimeter}\:\mathrm{150g}\:\mathrm{of}\:\mathrm{water}\:\mathrm{at}\:\mathrm{10}°\mathrm{C}. \\ $$$$\mathrm{The}\:\mathrm{temperature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mixture}\:\mathrm{is}\:\mathrm{45}°\mathrm{C}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{thermal}\:\mathrm{capaity}\:\mathrm{of}\:\mathrm{tbe}\:\mathrm{vessel}. \\ $$$$\mathrm{specific}\:\mathrm{heat}\:\mathrm{capacity}\:\mathrm{of}\:\mathrm{copper}\:=\:\mathrm{400} \\ $$$$\mathrm{specific}\:\mathrm{heat}\:\mathrm{capacity}\:\mathrm{of}\:\mathrm{water}\:=\:\mathrm{4200} \\ $$ Answered…