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Question Number 54896 Answers: 1 Comments: 1

∫(√)(x2+2x+2)

$$\int\sqrt{}\left({x}\mathrm{2}+\mathrm{2}{x}+\mathrm{2}\right) \\ $$

Question Number 54880 Answers: 2 Comments: 0

If x^2 −y^2 =a^2 find (d^2 y/dx^2 ) if a is constant.

$${If}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:{find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{if}\:{a}\:{is} \\ $$$${constant}. \\ $$

Question Number 54876 Answers: 1 Comments: 1

find the coefficientof x^2 in the binomial expansion of (x^2 +(2/x))^4

$$\mathrm{find}\:\mathrm{the}\:\mathrm{coefficientof}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{binomial} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{x}}\right)^{\mathrm{4}} \\ $$

Question Number 54875 Answers: 2 Comments: 0

Given that((log(3x+1)^(2x−1) )/(log(3x+1)))=5,find the value of x.

$$\mathrm{Given}\:\mathrm{that}\frac{\mathrm{log}\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2x}−\mathrm{1}} }{\mathrm{log}\left(\mathrm{3x}+\mathrm{1}\right)}=\mathrm{5},\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$

Question Number 54869 Answers: 1 Comments: 0

Question Number 54868 Answers: 0 Comments: 1

Question Number 54860 Answers: 1 Comments: 1

Question Number 54858 Answers: 2 Comments: 0

(Q1) The expansion (x−(1/(3x)))^(12) (i) find the term independent of x (ii) find the term x^4 (Q2) There are some nut in a bag. when 2 ounces of peanut are added to the mixture, the percentage of peanut becomes 20% . Richard added 2 ounces of cashew to the mixture and the percentage of cashew nut was 33.33% . find the percentage of cashew nut that were there initially. please sir help

$$\left({Q}\mathrm{1}\right)\:{The}\:{expansion}\:\left({x}−\frac{\mathrm{1}}{\mathrm{3}{x}}\right)^{\mathrm{12}} \\ $$$$\left({i}\right)\:{find}\:{the}\:{term}\:{independent}\:{of}\:{x} \\ $$$$\left({ii}\right)\:{find}\:{the}\:{term}\:{x}^{\mathrm{4}} \\ $$$$\left({Q}\mathrm{2}\right)\:{There}\:{are}\:{some}\:{nut}\:{in}\:{a}\:{bag}.\:{when} \\ $$$$\:\mathrm{2}\:{ounces}\:{of}\:{peanut}\:{are}\:{added}\:{to}\:{the} \\ $$$${mixture},\:{the}\:{percentage}\:{of}\:{peanut}\: \\ $$$${becomes}\:\mathrm{20\%}\:.\:{Richard}\:\:{added}\:\:\mathrm{2}\: \\ $$$${ounces}\:{of}\:{cashew}\:{to}\:{the}\:{mixture}\:{and} \\ $$$${the}\:{percentage}\:{of}\:{cashew}\:{nut}\:{was}\: \\ $$$$\mathrm{33}.\mathrm{33\%}\:.\:{find}\:{the}\:{percentage}\:{of}\: \\ $$$${cashew}\:{nut}\:{that}\:{were}\:{there}\:{initially}. \\ $$$${please}\:{sir}\:{help} \\ $$

Question Number 54857 Answers: 1 Comments: 0

Question Number 54856 Answers: 1 Comments: 2

Question Number 54841 Answers: 1 Comments: 0

Question Number 54830 Answers: 0 Comments: 1

let V_n = ∫_0 ^∞ ((cos(nx))/(n +x^2 ))dx with n integr nstural not 0 . 1) calculate V_n 2)calculate lim_(n→+∞) nV_n 3) calculate the sum Σ_(n=0) ^∞ V_n

$${let}\:{V}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}\:+{x}^{\mathrm{2}} }{dx}\:\:\:{with}\:{n}\:{integr}\:{nstural}\:{not}\:\mathrm{0}\:. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{V}_{{n}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} {nV}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{the}\:{sum}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{V}_{{n}} \\ $$

Question Number 54827 Answers: 0 Comments: 0

logx^2 +log_2 (x−6)=3 solve for x

$$\mathrm{logx}^{\mathrm{2}} +\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}−\mathrm{6}\right)=\mathrm{3} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$

Question Number 54826 Answers: 4 Comments: 5

1) ∫_0 ^( 1) ((1−x^7 )^(1/4) −(1−x^4 )^(1/7) )dx = ? 2) If f(x)=x^3 +3x+4 then the value of ∫_(−1) ^( 1) f(x)dx + ∫_0 ^( 4) f^( −1) (x)dx = ? 3) ∫_(−π) ^( π) (1+cosx+cos2x+....+cos13x)(1+sinx+...+sin13x)dx=? 4) ∫_0 ^( 2) ((√(x^3 +1)) + (x^2 +2x)^(1/3) )dx =? (This time func. are not inverse of each other,right ?)

$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} \right){dx}\:=\:? \\ $$$$\left.\mathrm{2}\right)\:{If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{4}\:{then}\:{the}\:{value}\:{of} \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} {f}\left({x}\right){dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{4}} {f}^{\:−\mathrm{1}} \left({x}\right){dx}\:=\:? \\ $$$$\left.\mathrm{3}\right)\:\int_{−\pi} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+....+{cos}\mathrm{13}{x}\right)\left(\mathrm{1}+{sinx}+...+{sin}\mathrm{13}{x}\right){dx}=? \\ $$$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\left(\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\right){dx}\:=? \\ $$$$\left({This}\:{time}\:{func}.\:{are}\:{not}\:{inverse}\:{of}\:{each}\right. \\ $$$$\left.{other},{right}\:?\right) \\ $$

Question Number 54825 Answers: 1 Comments: 0

Question Number 54808 Answers: 1 Comments: 2

let p(x)=(1+x^2 )(1+x^4 )...(1+x^2^n ) with n integr natural 1) find a simple form of p(x) 2) find roots of p(x)and decompose p(x) inside C[x] 3)calculate ∫_0 ^1 p(x)dx 4) decompose the fraction F(x)=(1/(p(x))) .

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)...\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$${with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{roots}\:{of}\:{p}\left({x}\right){and}\:{decompose} \\ $$$${p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{p}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{decompose}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)}\:. \\ $$

Question Number 54797 Answers: 1 Comments: 0

∫ (√( x + (√( x + (√(x + (√( .....)))))))) dx = ?

$$ \\ $$$$ \\ $$$$\:\:\:\int\:\:\sqrt{\:\boldsymbol{{x}}\:+\:\sqrt{\:\boldsymbol{{x}}\:+\:\sqrt{\boldsymbol{{x}}\:+\:\sqrt{\:.....}}}}\:\:\boldsymbol{{dx}}\:\:=\:\:\:? \\ $$$$ \\ $$

Question Number 54821 Answers: 0 Comments: 1

find lim_(n→+∞) ∫_0 ^n ((arctan(nx))/(n^2 +x^2 ))dx

$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{{arctan}\left({nx}\right)}{{n}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 54790 Answers: 3 Comments: 2

differenciatethefollowing i)x^x +(sinx)^(lnx) = ii)sin^(−1) (tanhx)= iii)(√(1+x^2 /1−x^2 =))

$${differenciatethefollowing} \\ $$$$\left.{i}\right){x}^{{x}} +\left({sinx}\right)^{{lnx}} = \\ $$$$\left.{ii}\right){sin}^{−\mathrm{1}} \left({tanhx}\right)= \\ $$$$\left.{iii}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} /\mathrm{1}−{x}^{\mathrm{2}} =} \\ $$

Question Number 54788 Answers: 2 Comments: 1

How can cut a right angeled triangle to make a square? how can cut a equilateral triangle to make a rectangle?

Question Number 54787 Answers: 0 Comments: 0

if: (a/((√b)+(√c)))=(b/((√c)+(√d)))=(c/((√d)+(√a)))=(d/((√a)+(√b))) ⇒(a/d)=?

$$\:{if}:\:\:\frac{{a}}{\sqrt{{b}}+\sqrt{{c}}}=\frac{{b}}{\sqrt{{c}}+\sqrt{{d}}}=\frac{{c}}{\sqrt{{d}}+\sqrt{{a}}}=\frac{{d}}{\sqrt{{a}}+\sqrt{{b}}} \\ $$$$\Rightarrow\frac{{a}}{{d}}=? \\ $$

Question Number 54786 Answers: 0 Comments: 5

solve for:a,b,c,d∈R a^2 =b+(√c) b^2 =c+(√d) c^2 =d+(√a) d^2 =a+(√b)

$${solve}\:{for}:{a},{b},{c},{d}\in\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} ={b}+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} ={c}+\sqrt{{d}} \\ $$$$\:\:\:\:\:\:{c}^{\mathrm{2}} ={d}+\sqrt{{a}} \\ $$$$\:\:\:\:\:\:{d}^{\mathrm{2}} ={a}+\sqrt{{b}} \\ $$

Question Number 54777 Answers: 0 Comments: 1

let u_n =∫_0 ^∞ ((sin(nx^2 ))/(x^2 +6))dx 1) calculate u_n and lim u_n (n→+∞) 2) find nature of Σ u_n and calaculate it. 3) find nature of Σ u_n ^2

$${let}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({nx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{6}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{u}_{{n}} \:\:\:{and}\:{lim}\:{u}_{{n}} \left({n}\rightarrow+\infty\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} \:\:\:{and}\:{calaculate}\:{it}. \\ $$$$\left.\mathrm{3}\right)\:{find}\:{nature}\:\:{of}\:\Sigma\:{u}_{{n}} ^{\mathrm{2}} \\ $$

Question Number 54775 Answers: 0 Comments: 2

found something interesting, it was published by Tschirnhaus in 1683 we can reduce x^3 +ax^2 +bx+c=0 (1) to y^3 +py+q=0 (2) and further to z^3 =t (1) is the well known linear substitution y=x+(a/3) → x=y−(a/3) ⇒ y^3 −((a^2 −3b)/3)y+((2a^3 −9ab+27c)/(27))=0 p=−((a^2 −3b)/3) and q=((2a^3 −9ab+27c)/(27)) ⇒ y^3 +py+q=0 (2) quadratic substitution z=y^2 +αy+β → y^2 +αy+(β−z)=0 we could solve this for y and then plug in above... (Tschirnhaus did) but there′s an easier way: we calculate the determinant of the Sylvester Matrix we have (a) 1y^3 +0y^2 +py+q=0 (b) 0y^3 +y^2 +αy+(β−z)=0 the matrix is [(1,0,p,q,0),(0,1,0,p,q),(0,1,α,(β−z),0),(0,0,1,α,(β−z)),(1,α,(β−z),0,0) ] the determinant is −z^3 +(3β−2p)z^2 −(pα^2 +3qα+3β^2 −4pβ+p^2 )z −(qα^3 −pα^2 β−3qαβ+pqα−β^3 +2pβ^2 −p^2 β−q^2 )p we want the square and the linear terms to disappear so we set their constants zero to get α and β ⇒ β=((2p)/3); α=−((3q)/(2p))±((√(12p^3 +81q^2 ))/(6p)) this leads to z^3 =((8p^3 )/(27))+((27q^4 )/(2p^3 ))+4q^2 ±((q(√(3(4p^3 +27q^2 )^3 )))/(18p^3 )) Tschirnhaus thought he could solve polynomes of any degree with this method but it′s getting harder to solve because you need a cubic substitution to eliminate 3 constants and so on...

$$\mathrm{found}\:\mathrm{something}\:\mathrm{interesting},\:\mathrm{it}\:\mathrm{was}\:\mathrm{published} \\ $$$$\mathrm{by}\:\mathrm{Tschirnhaus}\:\mathrm{in}\:\mathrm{1683} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{reduce} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{to} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{further}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} ={t} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{well}\:\mathrm{known}\:\mathrm{linear}\:\mathrm{substitution} \\ $$$${y}={x}+\frac{{a}}{\mathrm{3}}\:\rightarrow\:{x}={y}−\frac{{a}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} −\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}{y}+\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}}=\mathrm{0} \\ $$$${p}=−\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}\:\mathrm{and}\:{q}=\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{quadratic}\:\mathrm{substitution} \\ $$$${z}={y}^{\mathrm{2}} +\alpha{y}+\beta\:\rightarrow\:{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{y}\:\mathrm{and}\:\mathrm{then}\:\mathrm{plug}\:\mathrm{in} \\ $$$$\mathrm{above}...\:\left(\mathrm{Tschirnhaus}\:\mathrm{did}\right)\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{an} \\ $$$$\mathrm{easier}\:\mathrm{way}:\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{Sylvester}\:\mathrm{Matrix} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\left({a}\right)\:\mathrm{1}{y}^{\mathrm{3}} +\mathrm{0}{y}^{\mathrm{2}} +{py}+{q}=\mathrm{0} \\ $$$$\left({b}\right)\:\mathrm{0}{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{matrix}\:\mathrm{is} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}\\{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}\\{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\mathrm{the}\:\mathrm{determinant}\:\mathrm{is} \\ $$$$−{z}^{\mathrm{3}} \\ $$$$\:\:+\left(\mathrm{3}\beta−\mathrm{2}{p}\right){z}^{\mathrm{2}} \\ $$$$\:\:−\left({p}\alpha^{\mathrm{2}} +\mathrm{3}{q}\alpha+\mathrm{3}\beta^{\mathrm{2}} −\mathrm{4}{p}\beta+{p}^{\mathrm{2}} \right){z} \\ $$$$\:\:−\left({q}\alpha^{\mathrm{3}} −{p}\alpha^{\mathrm{2}} \beta−\mathrm{3}{q}\alpha\beta+{pq}\alpha−\beta^{\mathrm{3}} +\mathrm{2}{p}\beta^{\mathrm{2}} −{p}^{\mathrm{2}} \beta−{q}^{\mathrm{2}} \right){p} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{the}\:\mathrm{square}\:\mathrm{and}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{terms} \\ $$$$\mathrm{to}\:\mathrm{disappear}\:\mathrm{so}\:\mathrm{we}\:\mathrm{set}\:\mathrm{their}\:\mathrm{constants}\:\mathrm{zero} \\ $$$$\mathrm{to}\:\mathrm{get}\:\alpha\:\mathrm{and}\:\beta \\ $$$$\Rightarrow\:\beta=\frac{\mathrm{2}{p}}{\mathrm{3}};\:\alpha=−\frac{\mathrm{3}{q}}{\mathrm{2}{p}}\pm\frac{\sqrt{\mathrm{12}{p}^{\mathrm{3}} +\mathrm{81}{q}^{\mathrm{2}} }}{\mathrm{6}{p}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} =\frac{\mathrm{8}{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{27}{q}^{\mathrm{4}} }{\mathrm{2}{p}^{\mathrm{3}} }+\mathrm{4}{q}^{\mathrm{2}} \pm\frac{{q}\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{18}{p}^{\mathrm{3}} } \\ $$$$\mathrm{Tschirnhaus}\:\mathrm{thought}\:\mathrm{he}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{polynomes} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{degree}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\ $$$$\mathrm{harder}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{because}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{cubic} \\ $$$$\mathrm{substitution}\:\mathrm{to}\:\mathrm{eliminate}\:\mathrm{3}\:\mathrm{constants}\:\mathrm{and} \\ $$$$\mathrm{so}\:\mathrm{on}... \\ $$

Question Number 54770 Answers: 0 Comments: 0

A glass bottle full of mercury has mass 500g. On being heated through 35°C, 2.43g of mercury are expelled. Calculate the mass of mercury remaining in the bottle. (Cubic expansivity of mercury is 1.8 × 10^(−4) K^(−1) , linear expansivity of glass is 8.0 × 10^(−6) K^(−1) .

$$\mathrm{A}\:\mathrm{glass}\:\mathrm{bottle}\:\mathrm{full}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{has}\:\mathrm{mass}\:\mathrm{500g}.\:\mathrm{On}\:\mathrm{being}\:\mathrm{heated}\:\mathrm{through} \\ $$$$\mathrm{35}°\mathrm{C},\:\:\mathrm{2}.\mathrm{43g}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{are}\:\mathrm{expelled}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{mercury} \\ $$$$\mathrm{remaining}\:\mathrm{in}\:\mathrm{the}\:\mathrm{bottle}.\:\left(\mathrm{Cubic}\:\mathrm{expansivity}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{is}\:\:\mathrm{1}.\mathrm{8}\:×\:\mathrm{10}^{−\mathrm{4}} \:\mathrm{K}^{−\mathrm{1}} \:,\right. \\ $$$$\mathrm{linear}\:\mathrm{expansivity}\:\mathrm{of}\:\mathrm{glass}\:\mathrm{is}\:\:\mathrm{8}.\mathrm{0}\:×\:\mathrm{10}^{−\mathrm{6}} \mathrm{K}^{−\mathrm{1}} \:. \\ $$

Question Number 54767 Answers: 1 Comments: 1

A stone is thrown vertically upwards from a cliff 20m high. After a time of 3 s it passes the edge of the cliff on its way down. Calculate a) the speed of projection b) the speed when it hits the ground c) the times when it is 10m above the top of the cliff d) the time it is 15m above the ground (take g=10ms^(−2) ).

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