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In the figure below,a particle A of mass 2kg is lying on a rough wooden block.The particle A is connected by a light inextensible horizontal string passing over a smooth light fixed pulley at the edge of the block,to a particle B of mass 3kg which hangs freely. The coefficent of friction between the particle A and the surface of the block is μ. Given that the string is taut and the system is released from rest such that the particle move with an acceleration of 4ms^(−1) . Find a) the tension b) the value of μ.

$${In}\:{the}\:{figure}\:{below},{a}\:{particle}\:{A}\:{of} \\ $$$${mass}\:\mathrm{2}{kg}\:{is}\:{lying}\:{on}\:{a}\:{rough}\:{wooden} \\ $$$${block}.{The}\:{particle}\:{A}\:{is}\:{connected}\:{by} \\ $$$${a}\:{light}\:{inextensible}\:{horizontal}\:{string} \\ $$$${passing}\:{over}\:{a}\:{smooth}\:{light}\:{fixed} \\ $$$${pulley}\:{at}\:{the}\:{edge}\:{of}\:{the}\:{block},{to}\:{a} \\ $$$${particle}\:{B}\:{of}\:{mass}\:\mathrm{3}{kg}\:{which}\:{hangs} \\ $$$${freely}.\:{The}\:{coefficent}\:{of}\:{friction} \\ $$$${between}\:{the}\:{particle}\:{A}\:{and}\:{the}\:{surface} \\ $$$${of}\:{the}\:{block}\:{is}\:\mu. \\ $$$${Given}\:{that}\:{the}\:{string}\:{is}\:{taut}\:{and}\:{the}\:{system}\:{is}\:{released}\:{from}\:{rest}\:\:{such}\:{that}\:{the}\:{particle}\:{move}\:{with}\:{an} \\ $$$${acceleration}\:{of}\:\mathrm{4}{ms}^{−\mathrm{1}} .\:{Find} \\ $$$$\left.{a}\right)\:{the}\:{tension} \\ $$$$\left.{b}\right)\:{the}\:{value}\:{of}\:\mu. \\ $$

Question Number 38419 Answers: 1 Comments: 0

Question Number 38384 Answers: 0 Comments: 1

∫x^(−3cosx) dx please is this possible? If possibld then prove it. Thanks in advance.

$$\int{x}^{−\mathrm{3}{cosx}} {dx} \\ $$$$ \\ $$$${please}\:{is}\:{this}\:{possible}? \\ $$$${If}\:{possibld}\:{then}\:{prove}\:{it}. \\ $$$$ \\ $$$${Thanks}\:{in}\:{advance}. \\ $$

Question Number 38376 Answers: 1 Comments: 2

Question Number 38373 Answers: 1 Comments: 7

Question Number 38367 Answers: 2 Comments: 2

If f(x)=x^3 +1 then f^(−1) (x)=?

$${If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{1}\:{then}\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Question Number 38366 Answers: 1 Comments: 0

At time t,the force acting on a particle P of mass 2kg is (2ti + 4j)N.P is initially at rest at the point with position vector (i + 2j). Find: a) the velocity of P when t = 2. b) the position vector when t = 2.

$${At}\:{time}\:{t},{the}\:{force}\:{acting}\:{on}\:{a}\:{particle} \\ $$$${P}\:{of}\:{mass}\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}\boldsymbol{{ti}}\:+\:\mathrm{4}\boldsymbol{{j}}\right){N}.{P} \\ $$$${is}\:{initially}\:{at}\:{rest}\:{at}\:{the}\:{point}\:{with} \\ $$$${position}\:{vector}\:\left(\boldsymbol{{i}}\:+\:\mathrm{2}\boldsymbol{{j}}\right). \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}. \\ $$$$\left.{b}\right)\:{the}\:{position}\:{vector}\:{when}\:{t}\:=\:\mathrm{2}. \\ $$

Question Number 38362 Answers: 0 Comments: 0

Question Number 38365 Answers: 1 Comments: 0

A particle P moves on a straightline from a fixed point O and the distance x from O after t seconds is given as x = (1/(4 )) t^4 − (3/2) t^2 + 2t. Find: a) the velocity of P when t = 2, b) the acceleration of P when t = 2, c) the time at which the speed P is Minimum.

$${A}\:{particle}\:{P}\:{moves}\:{on}\:{a}\:{straightline} \\ $$$${from}\:{a}\:{fixed}\:{point}\:{O}\:{and}\:{the}\:{distance} \\ $$$${x}\:{from}\:{O}\:{after}\:{t}\:{seconds}\:{is}\:{given}\:{as} \\ $$$$\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}\:}\:{t}^{\mathrm{4}} \:−\:\frac{\mathrm{3}}{\mathrm{2}}\:{t}^{\mathrm{2}} \:+\:\mathrm{2}{t}. \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{b}\right)\:{the}\:{acceleration}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{c}\right)\:{the}\:{time}\:{at}\:{which}\:{the}\:{speed}\:{P}\: \\ $$$${is}\:{Minimum}. \\ $$

Question Number 38332 Answers: 1 Comments: 0

((3+(√5))/2) −(√((3+(√5))/2)) = ?!

$$\:\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\:?! \\ $$

Question Number 38310 Answers: 1 Comments: 4

let f(x)=∫_0 ^(+∞) ((arctan(xt))/(1+t^2 ))dt with x≥0 1) calculate f^′ (x) then a simple form of f(x) 2) calculate ∫_0 ^(+∞) ((arctant)/(1+t^2 ))dt 3) calculate ∫_0 ^(+∞) ((arctan(2t))/(1+t^2 ))dt

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{then}\:{a}\:{simple}\:{form}\:{of}\:\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$

Question Number 38323 Answers: 1 Comments: 1

find Σ_(n=1) ^(+∞) ((4n)/((2n−1)^2 (2n+1)^2 ))

$${find}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{4}{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 38296 Answers: 0 Comments: 7

i have a suggestion pls comment...we are all virtual friends common bond is mathematics so may know each other by posting our self photo...if administator give permission..

$${i}\:{have}\:{a}\:{suggestion}\:{pls}\:{comment}...{we}\:{are}\:{all} \\ $$$${virtual}\:{friends}\:{common}\:{bond}\:{is}\:{mathematics} \\ $$$${so}\:{may}\:{know}\:{each}\:{other}\:{by}\:{posting}\:{our}\:{self} \\ $$$${photo}...{if}\:{administator}\:{give}\:{permission}.. \\ $$

Question Number 38291 Answers: 1 Comments: 0

Question Number 38289 Answers: 1 Comments: 0

Question Number 38288 Answers: 1 Comments: 0

tan α−tan β = 2tan θ asin α−bsin β = lsin θ express sin α, sin β in terms of 𝛉.

$$\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta\:=\:\mathrm{2tan}\:\theta \\ $$$${a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\:{l}\mathrm{sin}\:\theta \\ $$$${express}\:\mathrm{sin}\:\alpha,\:\mathrm{sin}\:\beta\:\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}. \\ $$

Question Number 38286 Answers: 0 Comments: 1

(i) given the function f(t)=e^t and g(t)=lnt show that f○g(t)=g○f(t) (ii)if f(t)=at , g(t)=bt^2 +3 (fog)(2)=35 and (fog)(3)=75 find the value of a and b

$$\left(\boldsymbol{{i}}\right)\:\mathrm{given}\:\mathrm{the}\:\mathrm{function}\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\boldsymbol{\mathrm{e}}^{\boldsymbol{{t}}} \:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{\mathrm{ln}{t}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{{f}}\circ\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{g}}\circ\boldsymbol{{f}}\left(\boldsymbol{{t}}\right) \\ $$$$\left(\boldsymbol{{ii}}\right)\mathrm{if}\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{at}}\:,\:\boldsymbol{{g}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{bt}}^{\mathrm{2}} +\mathrm{3} \\ $$$$\left(\boldsymbol{{fog}}\right)\left(\mathrm{2}\right)=\mathrm{35}\:\boldsymbol{\mathrm{and}}\:\left(\boldsymbol{{fog}}\right)\left(\mathrm{3}\right)=\mathrm{75} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}} \\ $$

Question Number 38284 Answers: 1 Comments: 2

Find all the complex number in the rectangular form such that (z−1)^4 =−1

$${Find}\:{all}\:{the}\:{complex}\:{number}\:{in}\:{the} \\ $$$${rectangular}\:{form}\:{such}\:{that} \\ $$$$\left({z}−\mathrm{1}\right)^{\mathrm{4}} =−\mathrm{1} \\ $$$$ \\ $$

Question Number 38282 Answers: 1 Comments: 0

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