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Question Number 76809 Answers: 0 Comments: 3

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 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 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 76680 Answers: 0 Comments: 4

prove that: ∫_0 ^1 (1−x^7 )^(1/3) dx=∫_0 ^1 (1−x^3 )^(1/7) dx

$${prove}\:{that}: \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} {dx} \\ $$

Question Number 76579 Answers: 1 Comments: 0

A uniform ladder of weight W and length 2a rest in limiting equilibrium with one end on a rough horizontal ground and the other end on a rough vertical wall. The coefficient of friction between the ladder and the ground and between the ladder and the wall are respectively μ and λ . If the ladder makes an angle θ with the ground where tan θ = (5/(12)), a) show that 5μ + 6λμ − 6 = 0. b) find the value of λ and μ given that λμ =(1/2).

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{ladder}\:\mathrm{of}\:\mathrm{weight}\:{W}\:\mathrm{and}\:\mathrm{length} \\ $$$$\mathrm{2}{a}\:\mathrm{rest}\:\mathrm{in}\:\mathrm{limiting}\:\mathrm{equilibrium}\:\mathrm{with}\:\mathrm{one}\: \\ $$$$\mathrm{end}\:\mathrm{on}\:\mathrm{a}\:\mathrm{rough}\:\mathrm{horizontal}\:\mathrm{ground}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{end}\:\mathrm{on}\:\mathrm{a}\:\mathrm{rough}\:\mathrm{vertical}\:\mathrm{wall}. \\ $$$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{ladder} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{and}\:\mathrm{between}\:\mathrm{the}\:\mathrm{ladder}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{wall}\:\mathrm{are}\:\mathrm{respectively}\:\mu\:\mathrm{and}\:\lambda\:.\:\mathrm{If}\:\mathrm{the}\:\mathrm{ladder} \\ $$$$\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{where}\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{5}}{\mathrm{12}}, \\ $$$$\left.\mathrm{a}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{5}\mu\:+\:\mathrm{6}\lambda\mu\:−\:\mathrm{6}\:=\:\mathrm{0}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\lambda\:\mathrm{and}\:\mu\:\mathrm{given}\:\mathrm{that}\:\lambda\mu\:=\frac{\mathrm{1}}{\mathrm{2}}.\: \\ $$

Question Number 76577 Answers: 1 Comments: 0

given a sequence defined by {((3n)/(2n+ 5))}_(n=1) ^∞ , does this sequence converge or diverge, explain

$$\:{given}\:{a}\:{sequence}\:{defined}\:{by}\:\:\left\{\frac{\mathrm{3}{n}}{\mathrm{2}{n}+\:\mathrm{5}}\right\}_{{n}=\mathrm{1}} ^{\infty} ,\:{does}\:{this}\: \\ $$$${sequence}\:{converge}\:{or}\:{diverge},\:{explain} \\ $$

Question Number 76404 Answers: 0 Comments: 2

Hello have nice end of year good bless you all i respond note in y re message becsuse i have so many problemes that mack me feel no pleasur any more to do somthing i think its importante to say it i will back Soon i hop so Sorry for my English

$$\mathrm{Hello}\:\mathrm{have}\:\mathrm{nice}\:\mathrm{end}\:\mathrm{of}\:\mathrm{year}\:\mathrm{good}\:\mathrm{bless}\:\mathrm{you} \\ $$$$\mathrm{all}\:\mathrm{i}\:\mathrm{respond}\:\mathrm{note}\:\mathrm{in}\:\mathrm{y}\:\mathrm{re}\:\mathrm{message}\:\mathrm{becsuse}\:\mathrm{i}\:\mathrm{have}\:\mathrm{so}\:\mathrm{many}\:\mathrm{problemes} \\ $$$$\mathrm{that}\:\mathrm{mack}\:\mathrm{me}\:\mathrm{feel}\:\mathrm{no}\:\mathrm{pleasur}\:\mathrm{any}\:\mathrm{more}\:\mathrm{to}\:\mathrm{do}\:\mathrm{somthing} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{its}\:\mathrm{importante}\:\mathrm{to}\:\mathrm{say}\:\mathrm{it}\:\mathrm{i}\:\mathrm{will}\:\mathrm{back}\:\mathrm{Soon}\:\mathrm{i}\:\mathrm{hop}\:\mathrm{so}\:\:\mathrm{Sorry}\:\mathrm{for} \\ $$$$\mathrm{my}\:\mathrm{English} \\ $$

Question Number 76384 Answers: 0 Comments: 1

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Question Number 76368 Answers: 3 Comments: 5

prove that 1. Σ_(r=1) ^n r = (1/2)n(n+1) 2. Σ_(r=1) ^n r^2 = (1/6)n(n+1)(2n + 1) 3. Σ_(r=1) ^n r^3 = (1/4)n^2 (n + 1)^2

$${prove}\:{that} \\ $$$$\mathrm{1}.\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{2}.\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{6}}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\:+\:\mathrm{1}\right) \\ $$$$\mathrm{3}.\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}}{n}^{\mathrm{2}} \left({n}\:+\:\mathrm{1}\right)^{\mathrm{2}} \\ $$

Question Number 76367 Answers: 0 Comments: 4

prove that Σ_(r=1) ^∞ (1/r^2 ) = (π^2 /6)

$${prove}\:{that}\:\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Question Number 76229 Answers: 1 Comments: 0

Let P(x) be polynomial in x with integral coefficients. If n is a solution of P(x)≡0(mod n) , and a≡b(mod n), prove that b is also a solution.

$${Let}\:{P}\left({x}\right)\:{be}\:{polynomial}\:{in}\:{x}\:{with}\:{integral} \\ $$$${coefficients}.\:{If}\:{n}\:{is}\:{a}\:{solution}\:{of}\: \\ $$$${P}\left({x}\right)\equiv\mathrm{0}\left({mod}\:{n}\right)\:,\:{and}\:{a}\equiv{b}\left({mod}\:{n}\right), \\ $$$${prove}\:{that}\:{b}\:{is}\:{also}\:{a}\:{solution}. \\ $$