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Question Number 24057 Answers: 1 Comments: 0

A block of ice slides down a 45° incline plane in twice the time it takes to slide down a 45° frictionless incline plane.What is the coefficient of kinetic friction between the ice block and the incline plqne.

$${A}\:{block}\:{of}\:{ice}\:{slides}\:{down}\:{a}\:\mathrm{45}° \\ $$$${incline}\:{plane}\:{in}\:{twice}\:{the}\:{time}\:{it} \\ $$$${takes}\:{to}\:{slide}\:{down}\:{a}\:\mathrm{45}°\:{frictionless} \\ $$$${incline}\:{plane}.{What}\:{is}\:{the}\:{coefficient} \\ $$$${of}\:{kinetic}\:{friction}\:{between}\:{the} \\ $$$${ice}\:{block}\:{and}\:{the}\:{incline}\:{plqne}. \\ $$

Question Number 24055 Answers: 0 Comments: 2

Any Architect in the house? please i need your help

$${Any}\:{Architect}\:{in}\:{the}\:{house}? \\ $$$$ \\ $$$${please}\:{i}\:{need}\:{your}\:{help} \\ $$

Question Number 24054 Answers: 0 Comments: 0

Compare the bond strength of S − O bond in SO_3 ^(−2) and SO_4 ^(−2) ion.

$$\mathrm{Compare}\:\mathrm{the}\:\mathrm{bond}\:\mathrm{strength}\:\mathrm{of}\:\mathrm{S}\:−\:\mathrm{O} \\ $$$$\mathrm{bond}\:\mathrm{in}\:\mathrm{SO}_{\mathrm{3}} ^{−\mathrm{2}} \:\mathrm{and}\:\mathrm{SO}_{\mathrm{4}} ^{−\mathrm{2}} \:\mathrm{ion}. \\ $$

Question Number 24069 Answers: 1 Comments: 0

Simplify (((log_2 (√5) . log_(25) 20) + log_4 (√(50)) )/(log_4 70 − log_(15) 49))

$$\mathrm{Simplify} \\ $$$$\frac{\left(\mathrm{log}_{\mathrm{2}} \:\sqrt{\mathrm{5}}\:.\:\mathrm{log}_{\mathrm{25}} \:\mathrm{20}\right)\:+\:\mathrm{log}_{\mathrm{4}} \:\sqrt{\mathrm{50}}\:\:}{\mathrm{log}_{\mathrm{4}} \:\mathrm{70}\:−\:\mathrm{log}_{\mathrm{15}} \:\mathrm{49}} \\ $$

Question Number 24042 Answers: 1 Comments: 2

f(x) = sin(sin^2 x) + cos(sin^2 x) then the range of f(x) is

$${f}\left({x}\right)\:=\:\mathrm{sin}\left(\mathrm{sin}^{\mathrm{2}} {x}\right)\:+\:\mathrm{cos}\left(\mathrm{sin}^{\mathrm{2}} {x}\right)\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{is} \\ $$

Question Number 24037 Answers: 0 Comments: 0

Question Number 24032 Answers: 0 Comments: 7

A hemispherical bowl of radius R = 0.1 m is rotating about its own axis (which is vertical) with an angular velocity ω. A particle of mass 10^(−2) kg on the frictionless inner surface of the bowl is also rotating with the same ω. The particle is at a height h from the bottom of the bowl. It is desired to measure g (acceleration due to gravity) using the set up by measuring h accurately. Assuming that R and ω are known precisely and that the least count in the measurement of h is 10^(−4) m, what is the minimum possible error Δg in the measured value of g?

$$\mathrm{A}\:\mathrm{hemispherical}\:\mathrm{bowl}\:\mathrm{of}\:\mathrm{radius}\:{R}\:=\:\mathrm{0}.\mathrm{1} \\ $$$$\mathrm{m}\:\mathrm{is}\:\mathrm{rotating}\:\mathrm{about}\:\mathrm{its}\:\mathrm{own}\:\mathrm{axis}\:\left(\mathrm{which}\right. \\ $$$$\left.\mathrm{is}\:\mathrm{vertical}\right)\:\mathrm{with}\:\mathrm{an}\:\mathrm{angular}\:\mathrm{velocity}\:\omega. \\ $$$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{10}^{−\mathrm{2}} \:\mathrm{kg}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{frictionless}\:\mathrm{inner}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bowl}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{rotating}\:\mathrm{with}\:\mathrm{the}\:\mathrm{same}\:\omega.\:\mathrm{The} \\ $$$$\mathrm{particle}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:{h}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bottom} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{bowl}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{desired}\:\mathrm{to}\:\mathrm{measure}\:{g}\:\left(\mathrm{acceleration}\right. \\ $$$$\left.\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\right)\:\mathrm{using}\:\mathrm{the}\:\mathrm{set}\:\mathrm{up}\:\mathrm{by} \\ $$$$\mathrm{measuring}\:{h}\:\mathrm{accurately}.\:\mathrm{Assuming} \\ $$$$\mathrm{that}\:{R}\:\mathrm{and}\:\omega\:\mathrm{are}\:\mathrm{known}\:\mathrm{precisely}\:\mathrm{and} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{least}\:\mathrm{count}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measurement} \\ $$$$\mathrm{of}\:{h}\:\mathrm{is}\:\mathrm{10}^{−\mathrm{4}} \:\mathrm{m},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{possible}\:\mathrm{error}\:\Delta{g}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measured}\:\mathrm{value} \\ $$$$\mathrm{of}\:{g}? \\ $$

Question Number 24023 Answers: 0 Comments: 0

Compounds with high heat of formation are less stable because (1) it is difficult to synthesize them (2) energy rich state leads to instability (3) high temperature is required to synthesize them (4) molecules of such compounds are distorted

$$\mathrm{Compounds}\:\mathrm{with}\:\mathrm{high}\:\mathrm{heat}\:\mathrm{of}\:\mathrm{formation} \\ $$$$\mathrm{are}\:\mathrm{less}\:\mathrm{stable}\:\mathrm{because} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{it}\:\mathrm{is}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{synthesize}\:\mathrm{them} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{energy}\:\mathrm{rich}\:\mathrm{state}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{instability} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{high}\:\mathrm{temperature}\:\mathrm{is}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{synthesize}\:\mathrm{them} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{molecules}\:\mathrm{of}\:\mathrm{such}\:\mathrm{compounds}\:\mathrm{are} \\ $$$$\mathrm{distorted} \\ $$

Question Number 24022 Answers: 1 Comments: 0

A one kg ball rolling on a smooth horizontal surface at 20 m s^(−1) comes to the bottom of an inclined plane making an angle of 30° with the horizontal. Calculate K.E. of the ball when it is at the bottom of incline. How far up the incline will the ball roll? Neglect friction.

$$\mathrm{A}\:\mathrm{one}\:\mathrm{kg}\:\mathrm{ball}\:\mathrm{rolling}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{at}\:\mathrm{20}\:\mathrm{m}\:\mathrm{s}^{−\mathrm{1}} \:\mathrm{comes}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{bottom}\:\mathrm{of}\:\mathrm{an}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{making} \\ $$$$\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{30}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}. \\ $$$$\mathrm{Calculate}\:\mathrm{K}.\mathrm{E}.\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{bottom}\:\mathrm{of}\:\mathrm{incline}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{up}\:\mathrm{the} \\ $$$$\mathrm{incline}\:\mathrm{will}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{roll}?\:\mathrm{Neglect} \\ $$$$\mathrm{friction}. \\ $$

Question Number 24025 Answers: 1 Comments: 4

Question Number 23996 Answers: 1 Comments: 0

z^(−4_(=1/3(1−(√(3i)))) )

$$\mathrm{z}^{−\mathrm{4}_{=\mathrm{1}/\mathrm{3}\left(\mathrm{1}−\sqrt{\left.\mathrm{3i}\right)}\right.} } \\ $$

Question Number 24010 Answers: 0 Comments: 0

Prove that Σ_(r=1) ^(2n−1) (−1)^(r−1) ∙(r/(^(2n) C_r )) = (n/(n + 1)) .

$$\mathrm{Prove}\:\mathrm{that}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} \centerdot\frac{{r}}{\:^{\mathrm{2}{n}} {C}_{{r}} }\:=\:\frac{{n}}{{n}\:+\:\mathrm{1}}\:. \\ $$

Question Number 23984 Answers: 2 Comments: 0

Which of the following relation is/are correct? (1) ΔG = ΔH − TΔS (2) ΔG = ΔH + T[((δ(ΔG))/(δT))]_P (3) ΔG = ΔH + TΔS (4) ΔG = ΔH + ΔnRT

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relation}\:\mathrm{is}/\mathrm{are} \\ $$$$\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:−\:\mathrm{T}\Delta\mathrm{S} \\ $$$$\left(\mathrm{2}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\mathrm{T}\left[\frac{\delta\left(\Delta\mathrm{G}\right)}{\delta\mathrm{T}}\right]_{\mathrm{P}} \\ $$$$\left(\mathrm{3}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\mathrm{T}\Delta\mathrm{S} \\ $$$$\left(\mathrm{4}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\Delta\mathrm{nRT} \\ $$

Question Number 23960 Answers: 1 Comments: 4

Question Number 23957 Answers: 0 Comments: 0

∫((sinx−2cosx)/(1+2sin2x))dx solves

$$\int\frac{\boldsymbol{\mathrm{sinx}}−\mathrm{2}\boldsymbol{\mathrm{cosx}}}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{solves}} \\ $$

Question Number 23968 Answers: 0 Comments: 3

Prove that n and 2n−1 are coprime.

$${Prove}\:{that}\:{n}\:{and}\:\mathrm{2}{n}−\mathrm{1}\:{are}\:{coprime}. \\ $$

Question Number 23972 Answers: 0 Comments: 0

∫_0 ^(90) ((sin^2 x)/(sin^4 x+cos^4 x))dx

$$\int_{\mathrm{0}} ^{\mathrm{90}} \frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}{dx} \\ $$

Question Number 23953 Answers: 0 Comments: 4

find the nth term of the sequence 4, 9, 16, 45, 76 please help

$${find}\:{the}\:{nth}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\mathrm{4},\:\mathrm{9},\:\mathrm{16},\:\mathrm{45},\:\mathrm{76} \\ $$$$ \\ $$$${please}\:{help} \\ $$

Question Number 23940 Answers: 0 Comments: 2

When the gas is ideal and process is isothermal, then (1) P_1 V_1 = P_2 V_2 (2) ΔU = 0 (3) ΔW = 0 (4) ΔH_1 = ΔH_2

$$\mathrm{When}\:\mathrm{the}\:\mathrm{gas}\:\mathrm{is}\:\mathrm{ideal}\:\mathrm{and}\:\mathrm{process}\:\mathrm{is} \\ $$$$\mathrm{isothermal},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{P}_{\mathrm{1}} \mathrm{V}_{\mathrm{1}} \:=\:\mathrm{P}_{\mathrm{2}} \mathrm{V}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\Delta\mathrm{U}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\Delta\mathrm{W}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\Delta\mathrm{H}_{\mathrm{1}} \:=\:\Delta\mathrm{H}_{\mathrm{2}} \\ $$

Question Number 23939 Answers: 0 Comments: 0

In a school of 500 students,150 played hockey 250 played football,120 played table tennis and 100 played one of the three games.If 50 played both hockey and football,how many played table tennis only?pls show workings with diagram

$${In}\:{a}\:{school}\:{of}\:\mathrm{500}\:{students},\mathrm{150}\:{played}\:{hockey} \\ $$$$\mathrm{250}\:{played}\:{football},\mathrm{120}\:{played}\:{table}\:{tennis}\:{and} \\ $$$$\mathrm{100}\:{played}\:{one}\:{of}\:{the}\:{three}\:{games}.{If}\:\mathrm{50}\: \\ $$$${played}\:{both}\:{hockey}\:{and}\:{football},{how}\:{many} \\ $$$${played}\:{table}\:{tennis}\:{only}?{pls}\:{show}\:{workings}\:{with} \\ $$$${diagram} \\ $$$$ \\ $$

Question Number 23937 Answers: 1 Comments: 1

Question Number 23928 Answers: 0 Comments: 4

Find^n C_1 − (1/2)^n C_2 + (1/3)^n C_3 − .... + (−1)^(n−1) (1/n) ∙^n C_n .

$$\mathrm{Find}\:^{{n}} {C}_{\mathrm{1}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}\:^{{n}} {C}_{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{3}}\:^{{n}} {C}_{\mathrm{3}} \:−\:....\:+ \\ $$$$\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{n}}\:\centerdot\:^{{n}} {C}_{{n}} . \\ $$

Question Number 23932 Answers: 1 Comments: 0

A cord is wound around the circumference of a bicycle wheel (without tyre) of diameter 1 m. A mass of 2 kg is tied at the end of the cord and it is allowed to fall from rest. The weight falls 2 m in 4 s. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. If g = 10 ms^(−2) , what is the angular acceleration of the wheel?

$$\mathrm{A}\:\mathrm{cord}\:\mathrm{is}\:\mathrm{wound}\:\mathrm{around}\:\mathrm{the}\:\mathrm{circumference} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{bicycle}\:\mathrm{wheel}\:\left(\mathrm{without}\:\mathrm{tyre}\right)\:\mathrm{of} \\ $$$$\mathrm{diameter}\:\mathrm{1}\:\mathrm{m}.\:\mathrm{A}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{tied}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cord}\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{allowed}\:\mathrm{to} \\ $$$$\mathrm{fall}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{The}\:\mathrm{weight}\:\mathrm{falls}\:\mathrm{2}\:\mathrm{m}\:\mathrm{in}\:\mathrm{4} \\ $$$$\mathrm{s}.\:\mathrm{The}\:\mathrm{axle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wheel}\:\mathrm{is}\:\mathrm{horizontal} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{wheel}\:\mathrm{rotates}\:\mathrm{with}\:\mathrm{its}\:\mathrm{plane} \\ $$$$\mathrm{vertical}.\:\mathrm{If}\:\mathrm{g}\:=\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{2}} ,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{angular}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wheel}? \\ $$

Question Number 23922 Answers: 1 Comments: 0

if f(x)=2^x show that f(x+3)−f(x−1)=((15)/2)f(x)

$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}^{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\mathrm{3}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)=\frac{\mathrm{15}}{\mathrm{2}}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$

Question Number 23920 Answers: 1 Comments: 1

Question Number 23981 Answers: 1 Comments: 0

Amount of heat required to change 1 g ice at 0°C to 1 g steam at 100°C is (1) 616 cal (2) 12 kcal (3) 717 cal (4) none of these.

$$\mathrm{Amount}\:\mathrm{of}\:\mathrm{heat}\:\mathrm{required}\:\mathrm{to}\:\mathrm{change}\:\mathrm{1}\:{g} \\ $$$$\mathrm{ice}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}\:\mathrm{to}\:\mathrm{1}\:{g}\:\mathrm{steam}\:\mathrm{at}\:\mathrm{100}°\mathrm{C}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{616}\:\mathrm{cal} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{12}\:\mathrm{kcal} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{717}\:\mathrm{cal} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}. \\ $$

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