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AllQuestion and Answers: Page 1610

Question Number 40813    Answers: 0   Comments: 4

Question Number 40804    Answers: 0   Comments: 6

Question Number 40787    Answers: 1   Comments: 4

Let I_1 = ∫_(π/6) ^(π/3) ((sin x)/x) dx , I_2 = ∫_(π/6) ^(π/3) ((sin (sin x))/(sin x))dx , I_3 = ∫_(π/6) ^(π/3) ((sin (tan x))/(tan x))dx. Prove that I_2 > I_1 > I_3 .

$$\mathrm{Let}\:\mathrm{I}_{\mathrm{1}} =\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin}\:{x}}{{x}}\:{dx}\:\:,\:\:\mathrm{I}_{\mathrm{2}} =\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:{x}}{dx} \\ $$$$,\:\mathrm{I}_{\mathrm{3}} =\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{\mathrm{tan}\:{x}}{dx}.\: \\ $$$${P}\mathrm{rove}\:\mathrm{that}\:\mathrm{I}_{\mathrm{2}} \:>\:\mathrm{I}_{\mathrm{1}} \:>\:\mathrm{I}_{\mathrm{3}} \:. \\ $$

Question Number 40786    Answers: 1   Comments: 0

A parallel plate capacitor of plate spacing, 1mm is charged to a potential of 50V.Find the energy density in the capacitor

$${A}\:{parallel}\:{plate}\:{capacitor}\:{of}\:{plate} \\ $$$${spacing},\:\mathrm{1}{mm}\:{is}\:{charged}\:{to}\:{a}\:{potential} \\ $$$${of}\:\mathrm{50}{V}.{Find}\:{the}\:{energy}\:{density}\:{in} \\ $$$${the}\:{capacitor} \\ $$

Question Number 40782    Answers: 0   Comments: 0

let f(x)=ln(1+arctanx) 1) calculate f^((n)) (x) then f^((n)) (0) 2) developp f at integr serie .

$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{1}+{arctanx}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$

Question Number 40773    Answers: 1   Comments: 0

A pair of oppositely charged plane parallel plates each of area,100cm^2 has the electric field of the value 5.0×10^4 N/C.Calculate the charge on each plate.

$${A}\:{pair}\:{of}\:{oppositely}\:{charged}\:{plane} \\ $$$${parallel}\:{plates}\:{each}\:{of}\:{area},\mathrm{100}{cm}^{\mathrm{2}} \\ $$$${has}\:{the}\:{electric}\:{field}\:{of}\:{the}\:{value} \\ $$$$\mathrm{5}.\mathrm{0}×\mathrm{10}^{\mathrm{4}} {N}/{C}.{Calculate}\:{the}\:{charge} \\ $$$${on}\:{each}\:{plate}. \\ $$

Question Number 40771    Answers: 2   Comments: 1

Question Number 40770    Answers: 1   Comments: 0

An electric dipole is placed at rest in a uniform external electric field,and released.Discuss its motion mathematically.

$${An}\:{electric}\:{dipole}\:{is}\:{placed}\:{at}\:{rest} \\ $$$${in}\:{a}\:{uniform}\:{external}\:{electric} \\ $$$${field},{and}\:{released}.{Discuss}\:{its} \\ $$$${motion}\:{mathematically}. \\ $$

Question Number 40764    Answers: 1   Comments: 0

a piece of wire 40cm long is cut into two parts and each part is then bent into a square.if the sum of these squares is 68cm^2 find the lengths of the two pieces of wire.

$$\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{piece}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{wire}}\:\mathrm{40}\boldsymbol{\mathrm{cm}}\:\boldsymbol{\mathrm{long}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{parts}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{each}}\:\boldsymbol{\mathrm{part}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{bent}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{square}}.\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{these}}\:\boldsymbol{\mathrm{squares}}\:\boldsymbol{\mathrm{is}}\:\mathrm{68}\boldsymbol{\mathrm{cm}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{lengths}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{two}} \\ $$$$\boldsymbol{\mathrm{pieces}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{wire}}. \\ $$

Question Number 40763    Answers: 2   Comments: 0

Question Number 40760    Answers: 0   Comments: 0

find ∫ ((√(1+x^2 ))/(√(1−x^3 ))) dx

$${find}\:\:\int\:\:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}\:{dx} \\ $$

Question Number 40759    Answers: 1   Comments: 2

calculate lim_(n→+∞) ((1−e^(−nx^2 ) )/(x^2 sin((π/n))))

$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\frac{\mathrm{1}−{e}^{−{nx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} {sin}\left(\frac{\pi}{{n}}\right)} \\ $$

Question Number 40745    Answers: 3   Comments: 0

Question Number 40743    Answers: 2   Comments: 0

Question Number 40740    Answers: 0   Comments: 5

Question Number 40738    Answers: 2   Comments: 2

Question Number 40732    Answers: 1   Comments: 2

Question Number 40717    Answers: 1   Comments: 1

∫(√(tanx/sinx.cosxdx))

$$\int\sqrt{{tanx}/{sinx}.{cosxdx}} \\ $$

Question Number 40716    Answers: 2   Comments: 0

∫(cosx−cos2x/1−cosx)dx

$$\int\left({cosx}−{cos}\mathrm{2}{x}/\mathrm{1}−{cosx}\right){dx} \\ $$

Question Number 40715    Answers: 1   Comments: 1

for x≥2 ∣x−2∣=

$$\mathrm{for}\:\mathrm{x}\geqslant\mathrm{2}\:\mid\mathrm{x}−\mathrm{2}\mid= \\ $$

Question Number 40711    Answers: 1   Comments: 0

Two point charges,q_1 =0.4μC and q_2 =−0.3μC are placed at 10cm apart.Calculate (a)the potential at point A which is midway between them,and (b)point B which is 6cm from q_1 and 8cm from q_2

$${Two}\:{point}\:{charges},{q}_{\mathrm{1}} =\mathrm{0}.\mathrm{4}\mu{C}\:{and} \\ $$$${q}_{\mathrm{2}} =−\mathrm{0}.\mathrm{3}\mu{C}\:{are}\:{placed}\:{at}\:\mathrm{10}{cm} \\ $$$${apart}.{Calculate}\: \\ $$$$\left({a}\right){the}\:{potential}\:{at}\:{point}\:{A}\:{which}\:{is} \\ $$$${midway}\:{between}\:{them},{and} \\ $$$$\left({b}\right){point}\:{B}\:{which}\:{is}\:\mathrm{6}{cm}\:{from}\:{q}_{\mathrm{1}} \\ $$$${and}\:\mathrm{8}{cm}\:{from}\:{q}_{\mathrm{2}} \\ $$

Question Number 40709    Answers: 1   Comments: 1

calculate lim_(x→0) ((cos(x−sinx)−1)/(x^2 ))

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{cos}\left({x}−{sinx}\right)−\mathrm{1}}{{x}^{\mathrm{2}} \:} \\ $$

Question Number 40700    Answers: 1   Comments: 0

A charge q_1 =2.0×10^(−9) C is placed at the point,(x=0,y=4cm) and another charge,q_2 =−3.0×10^(−9) C is located at the point,(x=3cm,y=4cm). If a third charge,q_3 =4.0×10^(−9) C is placed at the origin, a)obtain the x and y component of the total force on q_3 b)Calculate the magnitude and direction of the total force on q_(3.)

$${A}\:{charge}\:{q}_{\mathrm{1}} =\mathrm{2}.\mathrm{0}×\mathrm{10}^{−\mathrm{9}} {C}\:{is}\:{placed} \\ $$$${at}\:{the}\:{point},\left({x}=\mathrm{0},{y}=\mathrm{4}{cm}\right)\:{and} \\ $$$${another}\:{charge},{q}_{\mathrm{2}} =−\mathrm{3}.\mathrm{0}×\mathrm{10}^{−\mathrm{9}} {C} \\ $$$${is}\:{located}\:{at}\:{the}\:{point},\left({x}=\mathrm{3}{cm},{y}=\mathrm{4}{cm}\right). \\ $$$${If}\:{a}\:{third}\:{charge},{q}_{\mathrm{3}} =\mathrm{4}.\mathrm{0}×\mathrm{10}^{−\mathrm{9}} {C} \\ $$$${is}\:{placed}\:{at}\:{the}\:{origin}, \\ $$$$\left.{a}\right){obtain}\:{the}\:{x}\:{and}\:{y}\:{component}\:{of} \\ $$$${the}\:{total}\:{force}\:{on}\:{q}_{\mathrm{3}} \\ $$$$\left.{b}\right){Calculate}\:{the}\:{magnitude}\:{and} \\ $$$${direction}\:{of}\:{the}\:{total}\:{force}\:{on}\:{q}_{\mathrm{3}.} \\ $$

Question Number 40699    Answers: 1   Comments: 0

Two point charges q_1 =1.5×10^(−9) C and q_2 =3.0×10^(−9) C are seperated by a distance of 200cm.Calculate the point at which the total electric field is zero.

$${Two}\:{point}\:{charges}\:{q}_{\mathrm{1}} =\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{9}} {C} \\ $$$${and}\:{q}_{\mathrm{2}} =\mathrm{3}.\mathrm{0}×\mathrm{10}^{−\mathrm{9}} {C}\:{are}\:{seperated} \\ $$$${by}\:{a}\:{distance}\:{of}\:\mathrm{200}{cm}.{Calculate} \\ $$$${the}\:{point}\:{at}\:{which}\:{the}\:{total}\:{electric} \\ $$$${field}\:{is}\:{zero}. \\ $$

Question Number 40686    Answers: 0   Comments: 6

Question Number 40684    Answers: 0   Comments: 1

∫((x^7 −1)/(logx))dx

$$\int\frac{\mathrm{x}^{\mathrm{7}} −\mathrm{1}}{\mathrm{logx}}\mathrm{dx} \\ $$

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