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Question Number 76802 Answers: 0 Comments: 0

3)αparticle of energy 5MeV pass through an ionisation chamber at the rate of 10 pwe second . Assum that all the energy is used in producing ion pairs,calculate the current produced. (35MeV is required for producing an ion pair and e=1.6×10^(-19) C) solution: Energy of α particles=5×10^6 eV Energy required for producing one ion pair=35eV No.of ion pairs produced by one α particle =((5×10^6 )/(35))=1.426×10^5 since 10 particle enter the chamber in one second =1.426×10^5 ×10=1.426×10^6 charge on dach ion=1.6×10^(-19) C Current=(1.426×10^6 )×(1.6×10^(-19) )C/s =2.287×10^(-13) A.

$$\left.\mathrm{3}\right)\alpha\mathrm{particle}\:\mathrm{of}\:\mathrm{energy}\:\mathrm{5MeV}\:\mathrm{pass} \\ $$$$\mathrm{through}\:\mathrm{an}\:\mathrm{ionisation}\:\mathrm{chamber}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{rate}\:\mathrm{of}\:\mathrm{10}\:\mathrm{pwe}\:\mathrm{second}\:.\:\mathrm{Assum}\:\mathrm{that}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{used}\:\mathrm{in}\:\mathrm{producing}\:\mathrm{ion}\: \\ $$$$\mathrm{pairs},\mathrm{calculate}\:\mathrm{the}\:\mathrm{current}\:\mathrm{produced}. \\ $$$$\left(\mathrm{35MeV}\:\mathrm{is}\:\mathrm{required}\:\mathrm{for}\:\mathrm{producing}\:\:\mathrm{an}\:\right. \\ $$$$\left.\mathrm{ion}\:\mathrm{pair}\:\mathrm{and}\:\mathrm{e}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \:\mathrm{C}\right) \\ $$$$\boldsymbol{\mathrm{solution}}: \\ $$$$\:\:\:\:\mathrm{Energy}\:\mathrm{of}\:\alpha\:\mathrm{particles}=\mathrm{5}×\mathrm{10}^{\mathrm{6}} \mathrm{eV} \\ $$$$\:\:\:\:\mathrm{Energy}\:\mathrm{required}\:\mathrm{for}\:\mathrm{producing}\:\mathrm{one} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{ion}\:\mathrm{pair}=\mathrm{35eV} \\ $$$$\:\:\:\:\mathrm{No}.\mathrm{of}\:\mathrm{ion}\:\mathrm{pairs}\:\mathrm{produced}\:\mathrm{by}\:\mathrm{one}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\mathrm{particle}\:=\frac{\mathrm{5}×\mathrm{10}^{\mathrm{6}} }{\mathrm{35}}=\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{5}} \\ $$$$\mathrm{since}\:\mathrm{10}\:\mathrm{particle}\:\mathrm{enter}\:\mathrm{the}\:\mathrm{chamber}\:\mathrm{in} \\ $$$$\mathrm{one}\:\mathrm{second} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{5}} ×\mathrm{10}=\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{6}} \\ $$$$\mathrm{charge}\:\mathrm{on}\:\mathrm{dach}\:\mathrm{ion}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \mathrm{C} \\ $$$$\mathrm{Current}=\left(\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{6}} \right)×\left(\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \right)\mathrm{C}/\mathrm{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}.\mathrm{287}×\mathrm{10}^{-\mathrm{13}} \mathrm{A}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 76794 Answers: 1 Comments: 1

Find the sum of x + (x/(1+x)) + (x/((1+x)^2 )) + (x/((1+x)^3 ))+.... for ∣(1/(1+x))∣<1

$${Find}\:{the}\:{sum}\:{of} \\ $$$${x}\:+\:\frac{{x}}{\mathrm{1}+{x}}\:+\:\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:+\:\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+....\:{for} \\ $$$$\mid\frac{\mathrm{1}}{\mathrm{1}+{x}}\mid<\mathrm{1} \\ $$

Question Number 76793 Answers: 2 Comments: 2

The sum of the first n terms of a series is given by: S_n =n^2 +7n+2. (i)Find a formula for the nth term (ii)write down the first 5 terms of the sequence

$${The}\:{sum}\:{of}\:{the}\:{first}\:{n}\:{terms}\:{of}\:{a}\:{series} \\ $$$${is}\:{given}\:{by}:\:{S}_{{n}} ={n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{2}. \\ $$$$\left({i}\right){Find}\:{a}\:{formula}\:{for}\:{the}\:{nth}\:{term} \\ $$$$\left({ii}\right){write}\:{down}\:{the}\:{first}\:\mathrm{5}\:{terms}\:{of}\:{the} \\ $$$${sequence} \\ $$$$ \\ $$

Question Number 76791 Answers: 0 Comments: 0

calculate ((sin44^o +sin66^o +sin70^o )/(cos22^o ×cos33^o ×cos35^o )).

$${calculate}\:\frac{{sin}\mathrm{44}^{{o}} +{sin}\mathrm{66}^{{o}} +{sin}\mathrm{70}^{{o}} }{{cos}\mathrm{22}^{{o}} ×{cos}\mathrm{33}^{{o}} ×{cos}\mathrm{35}^{{o}} }. \\ $$

Question Number 76790 Answers: 3 Comments: 1

calculate ((sin72^o +sin148^o +sin140^o )/(sin36^o ×sin74^o ×sin70^o )).

$${calculate}\:\frac{{sin}\mathrm{72}^{{o}} +{sin}\mathrm{148}^{{o}} +{sin}\mathrm{140}^{{o}} }{{sin}\mathrm{36}^{{o}} ×{sin}\mathrm{74}^{{o}} ×{sin}\mathrm{70}^{{o}} }. \\ $$

Question Number 76783 Answers: 0 Comments: 0

let f(x)=x^3 ,2π periodic odd developp f at fourier serie

$${let}\:{f}\left({x}\right)={x}^{\mathrm{3}} \:\:\:\:,\mathrm{2}\pi\:{periodic}\:{odd}\:{developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$

Question Number 76782 Answers: 1 Comments: 0

calculate ∫_0 ^∞ e^(−x) ((sin(x^2 ))/x^2 )dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} \:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 76781 Answers: 0 Comments: 2

find ∫_0 ^∞ ((1−e^(−x^2 ) )/x^2 )dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 76780 Answers: 0 Comments: 1

find A=∫_(−∞) ^(+∞) x e^(−x^2 ) arctan(x−(1/x))dx

$${find}\:{A}=\int_{−\infty} ^{+\infty} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$

Question Number 76779 Answers: 0 Comments: 1

calculate ∫_0 ^π ((x^2 cosx)/(3+sin^2 x))dx

$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}^{\mathrm{2}} {cosx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$

Question Number 76777 Answers: 0 Comments: 0

prove that ((cos(40°)))^(1/3) + ((cos(80°)))^(1/3) − ((cos(20°)))^(1/3) =(((3/2)((9)^(1/3) −2)))^(1/3)

$${prove}\:{that} \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{{cos}\left(\mathrm{40}°\right)}\:+\:\sqrt[{\mathrm{3}}]{{cos}\left(\mathrm{80}°\right)}\:−\:\sqrt[{\mathrm{3}}]{{cos}\left(\mathrm{20}°\right)}\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}}{\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{2}\right)} \\ $$

Question Number 76774 Answers: 0 Comments: 6

The probability that the birth days of six different persons will fall in exactly two calendar months is

$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{birth}\:\mathrm{days}\:\mathrm{of} \\ $$$$\mathrm{six}\:\mathrm{different}\:\mathrm{persons}\:\mathrm{will}\:\mathrm{fall}\:\mathrm{in}\:\mathrm{exactly} \\ $$$$\mathrm{two}\:\mathrm{calendar}\:\mathrm{months}\:\mathrm{is} \\ $$

Question Number 76772 Answers: 0 Comments: 0

Question Number 76771 Answers: 0 Comments: 0

Question Number 76770 Answers: 0 Comments: 0

Question Number 76740 Answers: 0 Comments: 8

15 persons, among whom are A and B, sit down at random at a round table. The probability that there are 4 persons between A and B is

$$\mathrm{15}\:\mathrm{persons},\:\mathrm{among}\:\mathrm{whom}\:\mathrm{are}\:{A}\:\mathrm{and}\:{B}, \\ $$$$\mathrm{sit}\:\mathrm{down}\:\mathrm{at}\:\mathrm{random}\:\mathrm{at}\:\mathrm{a}\:\mathrm{round}\:\mathrm{table}. \\ $$$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{persons} \\ $$$$\mathrm{between}\:{A}\:\mathrm{and}\:{B}\:\mathrm{is} \\ $$

Question Number 76726 Answers: 3 Comments: 0

var(x) = 2 then var(2x −3)=? E(x) = 2 then E(2x −3) = ?

$$\:\mathrm{var}\left(\mathrm{x}\right)\:=\:\mathrm{2}\:\mathrm{then}\:\mathrm{var}\left(\mathrm{2x}\:−\mathrm{3}\right)=? \\ $$$$\mathrm{E}\left(\mathrm{x}\right)\:=\:\mathrm{2}\:\mathrm{then}\:\mathrm{E}\left(\mathrm{2x}\:−\mathrm{3}\right)\:=\:? \\ $$

Question Number 76725 Answers: 0 Comments: 1

To prove that x^2 >y^2 , it is sufficient to prove that A. x > y B. x^3 >y^3 C. ∣x∣ > ∣y∣ D. x > 3y

$$\mathrm{To}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{x}^{\mathrm{2}} >\mathrm{y}^{\mathrm{2}} ,\:\mathrm{it}\:\mathrm{is}\:\mathrm{sufficient}\:\mathrm{to}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{A}.\:\mathrm{x}\:>\:\mathrm{y} \\ $$$$\mathrm{B}.\:\mathrm{x}^{\mathrm{3}} >\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{C}.\:\mid\mathrm{x}\mid\:>\:\mid\mathrm{y}\mid \\ $$$$\mathrm{D}.\:\mathrm{x}\:>\:\mathrm{3y} \\ $$

Question Number 76724 Answers: 1 Comments: 1

∫_0 ^3 ∣x^2 −1∣ dx ≡

$$\int_{\mathrm{0}} ^{\mathrm{3}} \mid\mathrm{x}^{\mathrm{2}} −\mathrm{1}\mid\:\mathrm{dx}\:\equiv\: \\ $$

Question Number 76723 Answers: 0 Comments: 0

the maclaurin expansion of ln (3 + 4x) is valid for A) −(3/4) ≤ x< (3/4) B) −(3/4)< x ≤ (3/4) C) −(1/4)< x ≤ (1/4) D) −(3/4)< x < (3/4)

$$\mathrm{the}\:\mathrm{maclaurin}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{ln}\:\left(\mathrm{3}\:+\:\mathrm{4}{x}\right)\:{is}\:{valid}\:{for} \\ $$$$\left.{A}\right)\:\:−\frac{\mathrm{3}}{\mathrm{4}}\:\leqslant\:\mathrm{x}<\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left.\mathrm{B}\right)\:−\frac{\mathrm{3}}{\mathrm{4}}<\:\mathrm{x}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left.\mathrm{C}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}<\:\mathrm{x}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{D}\right)\:−\frac{\mathrm{3}}{\mathrm{4}}<\:\mathrm{x}\:<\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Question Number 76721 Answers: 2 Comments: 0

what is the mean value of (1/(1+4x^(2 ) )) for 0≤x≤(1/2)

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4x}^{\mathrm{2}\:} }\:\:\mathrm{for}\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 76718 Answers: 0 Comments: 0

For n ∈ N prove by mathematical induction that cos α+cos (α+β)+cos [α+(n−1)β]+...cos [α+(n−1)β]= ((cos [α+(((n−1)/2))β]sin ((nβ)/2))/(sin (n/2)))

$${For}\:{n}\:\in\:{N}\:{prove}\:{by}\:{mathematical} \\ $$$${induction}\:{that} \\ $$$$\mathrm{cos}\:\alpha+\mathrm{cos}\:\left(\alpha+\beta\right)+\mathrm{cos}\:\left[\alpha+\left({n}−\mathrm{1}\right)\beta\right]+...\mathrm{cos}\:\left[\alpha+\left({n}−\mathrm{1}\right)\beta\right]= \\ $$$$\frac{\mathrm{cos}\:\left[\alpha+\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)\beta\right]\mathrm{sin}\:\frac{{n}\beta}{\mathrm{2}}}{\mathrm{sin}\:\frac{{n}}{\mathrm{2}}} \\ $$

Question Number 76717 Answers: 1 Comments: 0

A triangle is formed by the three straight line y=m_1 x+(a/m_1 ) y=m_2 x+(a/m_2 ) y=m_3 x+(a/m_3 ) prove that its orthocenter always lies on the line x+a=0

$${A}\:{triangle}\:{is}\:{formed}\:{by} \\ $$$${the}\:{three}\:{straight}\:{line} \\ $$$${y}={m}_{\mathrm{1}} {x}+\frac{{a}}{{m}_{\mathrm{1}} } \\ $$$${y}={m}_{\mathrm{2}} {x}+\frac{{a}}{{m}_{\mathrm{2}} } \\ $$$${y}={m}_{\mathrm{3}} {x}+\frac{{a}}{{m}_{\mathrm{3}} } \\ $$$${prove}\:{that}\:{its}\:{orthocenter} \\ $$$${always}\:{lies}\:{on}\:{the}\:{line} \\ $$$${x}+{a}=\mathrm{0} \\ $$$$ \\ $$

Question Number 76716 Answers: 1 Comments: 4

prove that ∫_0 ^π ((xsin x)/(1+cos^2 x))=(π^2 /4)

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$

Question Number 76715 Answers: 0 Comments: 3

If u=arcsin (x/y)+arctan (y/x) show that x(∂u/dx)+y(∂u/dy)=0

$${If}\:{u}={arc}\mathrm{sin}\:\frac{{x}}{{y}}+{arc}\mathrm{tan}\:\frac{{y}}{{x}} \\ $$$${show}\:{that}\: \\ $$$${x}\frac{\partial{u}}{{dx}}+{y}\frac{\partial{u}}{{dy}}=\mathrm{0} \\ $$

Question Number 76714 Answers: 1 Comments: 0

If y=(√(tan x+(√(tan x+(√(tan x+....∞)))))) prove that (dy/dx)=((sec^2 x)/(2y−1))

$${If}\:{y}=\sqrt{\mathrm{tan}\:{x}+\sqrt{\mathrm{tan}\:{x}+\sqrt{\mathrm{tan}\:{x}+....\infty}}}\: \\ $$$${prove}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{sec}\:^{\mathrm{2}} {x}}{\mathrm{2}{y}−\mathrm{1}} \\ $$

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