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m-1-s-1-x-m-2-s-2-y-x-y-solve-it-as-an-equation-

Question Number 27536 by darkalestero1@gmail.com last updated on 08/Jan/18 $$\boldsymbol{\mathrm{m}}_{\mathrm{1}} \boldsymbol{\mathrm{s}}_{\mathrm{1}} \left(\boldsymbol{\mathrm{x}}−\boldsymbol{\theta}\right)=\boldsymbol{\mathrm{m}}_{\mathrm{2}} \boldsymbol{\mathrm{s}}_{\mathrm{2}} \left(\boldsymbol{\theta}−\boldsymbol{\mathrm{y}}\right)\:\:\:;\:\boldsymbol{\mathrm{x}}=?\:;\boldsymbol{\mathrm{y}}=?\:\boldsymbol{\theta}=? \\ $$$$\mathrm{solve}\:\mathrm{it}\:\mathrm{as}\:\mathrm{an}\:\mathrm{equation}…. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

is-8-9-is-the-multiplicative-inerse-of-1-1-8-why-or-why-not-

Question Number 93044 by fath035990 last updated on 10/May/20 $$\mathrm{is}\:\frac{\mathrm{8}}{\mathrm{9}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{multiplicative}\:\mathrm{inerse}\:\mathrm{of}\:−\mathrm{1}\:\frac{\mathrm{1}}{\mathrm{8}}? \\ $$$$\mathrm{why}\:\mathrm{or}\:\mathrm{why}\:\mathrm{not}? \\ $$ Commented by prakash jain last updated on 10/May/20 $$−\mathrm{1}\frac{\mathrm{1}}{\mathrm{8}}=−\frac{\mathrm{9}}{\mathrm{8}} \\ $$$$\frac{\mathrm{8}}{\mathrm{9}}×\left(−\frac{\mathrm{9}}{\mathrm{8}}\right)=−\mathrm{1}…

A-2000kg-space-capsule-is-traveling-away-from-the-earth-determine-the-gravitational-field-strenght-and-gravitational-force-on-the-capsule-due-to-the-earth-when-it-is-a-At-a-distance-from-the-earth-

Question Number 27430 by tawa tawa last updated on 06/Jan/18 $$\mathrm{A}\:\mathrm{2000kg}\:\mathrm{space}\:\mathrm{capsule}\:\mathrm{is}\:\mathrm{traveling}\:\mathrm{away}\:\mathrm{from}\:\mathrm{the}\:\mathrm{earth},\:\mathrm{determine}\:\mathrm{the} \\ $$$$\mathrm{gravitational}\:\mathrm{field}\:\mathrm{strenght}\:\mathrm{and}\:\mathrm{gravitational}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the}\:\mathrm{capsule}\:\mathrm{due}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{earth}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{At}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{earth}'\mathrm{s}\:\mathrm{surface}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{At}\:\mathrm{a}\:\mathrm{very}\:\mathrm{large}\:\mathrm{distance}\:\mathrm{away}\:\mathrm{from}\:\mathrm{the}\:\mathrm{earth}\:\left(\mathrm{Take}\:\:\mathrm{g}\:=\:\mathrm{9}.\mathrm{8Nkg}^{−\mathrm{1}} \:\mathrm{on}\right. \\ $$$$\left.\mathrm{earth}\:\mathrm{surface}\right) \\ $$ Terms…

Find-the-workdone-in-moving-an-object-along-a-vector-r-3i-2j-5k-if-the-applied-force-is-F-2i-j-k-

Question Number 27428 by tawa tawa last updated on 06/Jan/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{workdone}\:\mathrm{in}\:\mathrm{moving}\:\mathrm{an}\:\mathrm{object}\:\mathrm{along}\:\mathrm{a}\:\mathrm{vector} \\ $$$$\mathrm{r}\:=\:\mathrm{3i}\:+\:\mathrm{2j}\:−\:\mathrm{5k}\:\:\:\mathrm{if}\:\mathrm{the}\:\mathrm{applied}\:\mathrm{force}\:\mathrm{is}\:\:\mathrm{F}\:=\:\mathrm{2i}\:−\:\mathrm{j}\:−\:\mathrm{k} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

A-particle-of-mass-2kg-moves-in-a-force-field-depending-on-a-time-t-given-by-F-24t-2-i-36t-16-j-12tk-assuming-that-at-t-0-the-particle-is-located-at-r-0-3i-j-4k-and-has-v-0-

Question Number 27427 by tawa tawa last updated on 06/Jan/18 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2kg}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{force}\:\mathrm{field}\:\mathrm{depending}\:\mathrm{on}\:\mathrm{a}\:\mathrm{time}\:\mathrm{t}\:\mathrm{given}\:\mathrm{by} \\ $$$$\mathrm{F}\:=\:\mathrm{24t}^{\mathrm{2}} \mathrm{i}\:+\:\left(\mathrm{36t}\:−\:\mathrm{16}\right)\mathrm{j}\:−\:\mathrm{12tk}\:\:\:\mathrm{assuming}\:\mathrm{that}\:\mathrm{at}\:\mathrm{t}\:=\:\mathrm{0}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{located} \\ $$$$\mathrm{at}\:\:\mathrm{r}_{\mathrm{0}} \:=\:\mathrm{3i}\:−\:\mathrm{j}\:+\:\mathrm{4k}\:\:\mathrm{and}\:\:\mathrm{has}\:\:\:\mathrm{v}_{\mathrm{0}} \:=\:\mathrm{6i}\:+\:\mathrm{5j}\:−\:\mathrm{8k}.\:\mathrm{Find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Velocity}\:\mathrm{at}\:\mathrm{any}\:\mathrm{time}\:\mathrm{t} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Position}\:\mathrm{at}\:\mathrm{any}\:\mathrm{time}\:\mathrm{t} \\ $$$$\left(\mathrm{c}\right)\:\tau\:\left(\mathrm{torgue}\right)\:\mathrm{at}\:\mathrm{any}\:\mathrm{time}\:\mathrm{t} \\…

I-n-0-1-x-2n-1-1-x-2-dx-n-0-prove-that-n-0-2n-1-I-n-2-2nI-n-1-

Question Number 158477 by LEKOUMA last updated on 04/Nov/21 $${I}_{{n}\:} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:,\:{n}\geqslant\mathrm{0}\: \\ $$$${prove}\:{that}\:\forall\:{n}\geqslant\mathrm{0}\: \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right){I}_{{n}} =\sqrt{\mathrm{2}}−\mathrm{2}{nI}_{{n}−\mathrm{1}} \\ $$ Answered by Ar…

A-and-B-are-walking-along-a-circular-track-They-start-from-same-point-at-8-00-am-A-can-walk-2-rounds-per-hour-and-B-can-walk-3-rounds-per-hour-How-many-times-they-cross-each-other-before-9-30-am-if-

Question Number 27399 by Rasheed.Sindhi last updated on 06/Jan/18 $$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{walking}\:\mathrm{along}\:\mathrm{a} \\ $$$$\mathrm{circular}\:\mathrm{track}.\mathrm{They}\:\mathrm{start}\:\mathrm{from} \\ $$$$\mathrm{same}\:\mathrm{point}\:\mathrm{at}\:\mathrm{8}:\mathrm{00}\:\mathrm{am}. \\ $$$$\mathrm{A}\:\mathrm{can}\:\mathrm{walk}\:\mathrm{2}\:\mathrm{rounds}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{and}\:\mathrm{B}\:\mathrm{can}\:\mathrm{walk}\:\mathrm{3}\:\mathrm{rounds}\:\mathrm{per}\:\mathrm{hour}. \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{times}\:\mathrm{they}\:\mathrm{cross}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{before}\:\mathrm{9}:\mathrm{30}\:\mathrm{am}\:\mathrm{if}\:\mathrm{they}\:\mathrm{walk} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Opposite}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\…

dt-3sint-4cost-

Question Number 92910 by s.ayeni14@yahoo.com last updated on 09/May/20 $$\int\frac{\mathrm{dt}}{\mathrm{3sint}+\mathrm{4cost}} \\ $$ Commented by msup by abdo last updated on 09/May/20 $${I}=\int\:\frac{{dt}}{\mathrm{3}{sint}\:+\mathrm{4}{cost}}\:{vhangement} \\ $$$${tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\:{give} \\…