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AllQuestion and Answers: Page 1833

Question Number 16579 Answers: 0 Comments: 3

Question Number 16570 Answers: 0 Comments: 1

A,B and E are three circles all with radius 1 unit.A and E touch at P whole B and E touch at Q. ∠POQ=x° where O is the centre of E.Find the area of the overlapping portion of A and B if 0≤x≤60°

$$\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{E}\:\mathrm{are}\:\mathrm{three}\:\mathrm{circles}\:\mathrm{all}\: \\ $$$$\mathrm{with}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{unit}.\mathrm{A}\:\mathrm{and}\:\mathrm{E}\:\mathrm{touch} \\ $$$$\mathrm{at}\:\mathrm{P}\:\mathrm{whole}\:\mathrm{B}\:\mathrm{and}\:\mathrm{E}\:\mathrm{touch}\:\mathrm{at}\:\mathrm{Q}. \\ $$$$\angle\mathrm{POQ}=\mathrm{x}°\:\mathrm{where}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centre}\: \\ $$$$\mathrm{of}\:\mathrm{E}.\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{if} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{60}° \\ $$

Question Number 16572 Answers: 1 Comments: 0

Question Number 16542 Answers: 0 Comments: 1

if (a/(∣Z_2 −Z_3 ∣))=(b/(∣Z_3 −Z_1 ∣))=(c/(∣Z_1 −Z_2 ∣)) Then.. find.. (a^2 /(Z_2 −Z_3 )) + (b^2 /(Z_3 −Z_1 )) + (c^2 /(Z_1 −Z_2 )) a) 0 b) 1 c) 2 d)N.O.T

$${if}\:\frac{{a}}{\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid}=\frac{{b}}{\mid{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \mid}=\frac{{c}}{\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid}\:\:{Then}.. \\ $$$${find}..\:\frac{{a}^{\mathrm{2}} }{{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} }\:+\:\frac{{b}^{\mathrm{2}} }{{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} }\:+\:\frac{{c}^{\mathrm{2}} }{{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} }\: \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{0} \\ $$$$\left.{b}\right)\:\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{2} \\ $$$$\left.{d}\right){N}.{O}.{T} \\ $$

Question Number 16540 Answers: 2 Comments: 0

if ∣Z∣=1 Then ((1+Z)/(1+Z^ )) is equal to... a) Z b) Z^ c) Z+Z^ d) N.O.T

$${if}\:\mid{Z}\mid=\mathrm{1}\:{Then}\:\frac{\mathrm{1}+{Z}}{\mathrm{1}+\bar {{Z}}}\:\:{is}\:{equal}\:{to}... \\ $$$$\left.{a}\right)\:{Z}\:\: \\ $$$$\left.{b}\right)\:\:\bar {{Z}} \\ $$$$\left.{c}\right)\:{Z}+\bar {{Z}} \\ $$$$\left.{d}\right)\:{N}.{O}.{T} \\ $$

Question Number 16547 Answers: 1 Comments: 2

solve dy/dx + xy=y^2 e^(1/2x^2 ) logx

$${solve} \\ $$$${dy}/{dx}\:+\:{xy}={y}^{\mathrm{2}} {e}^{\mathrm{1}/\mathrm{2}{x}^{\mathrm{2}} } {logx} \\ $$

Question Number 16546 Answers: 2 Comments: 0

dv/dt=−(kv+bt) where k and b are constants solve the equation of v given v=u when t=0

$${dv}/{dt}=−\left({kv}+{bt}\right)\:{where}\:{k}\:{and}\:{b}\:{are}\:{constants} \\ $$$${solve}\:{the}\:{equation}\:{of}\:{v}\:{given} \\ $$$${v}={u}\:{when}\:{t}=\mathrm{0} \\ $$

Question Number 16528 Answers: 1 Comments: 0

Question Number 16519 Answers: 2 Comments: 0

Solve: ∣2 − x∣ − 2 ∣x + 1∣ < 1

$$\mathrm{Solve}: \\ $$$$\mid\mathrm{2}\:−\:\mathrm{x}\mid\:−\:\mathrm{2}\:\mid\mathrm{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{1} \\ $$

Question Number 16514 Answers: 1 Comments: 1

Help plz Given 2d^2 x/dt^2 +2dx/dt+x=ksint where k is constant show that if x=n and dx/dt=0 when t=0 then x=e^(−1/2t) {(n+2/5k)cos(t/2)+)+(n+4/5k)sin(t/2)} −1/5ksint−2/5kcost.

$${Help}\:{plz} \\ $$$${Given}\:\mathrm{2}{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +\mathrm{2}{dx}/{dt}+{x}={ksint} \\ $$$${where}\:{k}\:{is}\:{constant}\:{show}\:{that}\:{if}\:{x}={n} \\ $$$${and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0}\:{then}\: \\ $$$$\left.{x}={e}^{−\mathrm{1}/\mathrm{2}{t}} \left\{\left({n}+\mathrm{2}/\mathrm{5}{k}\right){cos}\left({t}/\mathrm{2}\right)+\right)+\left({n}+\mathrm{4}/\mathrm{5}{k}\right){sin}\left({t}/\mathrm{2}\right)\right\} \\ $$$$−\mathrm{1}/\mathrm{5}{ksint}−\mathrm{2}/\mathrm{5}{kcost}. \\ $$

Question Number 16506 Answers: 0 Comments: 0

A train 1 moves from east to west (clockwise) and another train 2 moves from west to east (anticlockwise) on the equator with equal speeds relative to ground. The ratio of their centripetal acceleration (a_1 /a_2 ) relative to centre of earth is (1) > 1 (2) < 1

$$\mathrm{A}\:\mathrm{train}\:\mathrm{1}\:\mathrm{moves}\:\mathrm{from}\:\mathrm{east}\:\mathrm{to}\:\mathrm{west} \\ $$$$\left(\mathrm{clockwise}\right)\:\mathrm{and}\:\mathrm{another}\:\mathrm{train}\:\mathrm{2}\:\mathrm{moves} \\ $$$$\mathrm{from}\:\mathrm{west}\:\mathrm{to}\:\mathrm{east}\:\left(\mathrm{anticlockwise}\right)\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{equator}\:\mathrm{with}\:\mathrm{equal}\:\mathrm{speeds}\:\mathrm{relative} \\ $$$$\mathrm{to}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{their} \\ $$$$\mathrm{centripetal}\:\mathrm{acceleration}\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }\:\mathrm{relative}\:\mathrm{to} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{earth}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:<\:\mathrm{1} \\ $$

Question Number 16505 Answers: 1 Comments: 0

What is the maximum magnitude of change in velocity of a motorcycle moving with a uniform speed v_0 in a circular path of length l = (π/3)R and radius R? Treat motorcycle as a particle (1) ∣Δv^→ ∣ = (v_0 /2) (2) ∣Δv^→ ∣ = v_0

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{magnitude}\:\mathrm{of} \\ $$$$\mathrm{change}\:\mathrm{in}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{motorcycle} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circular}\:\mathrm{path}\:\mathrm{of}\:\mathrm{length}\:{l}\:=\:\frac{\pi}{\mathrm{3}}{R}\:\mathrm{and} \\ $$$$\mathrm{radius}\:\mathrm{R}?\:\mathrm{Treat}\:\mathrm{motorcycle}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{particle} \\ $$$$\left(\mathrm{1}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:\frac{{v}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mid\Delta\overset{\rightarrow} {{v}}\mid\:=\:{v}_{\mathrm{0}} \\ $$

Question Number 16504 Answers: 2 Comments: 0

An object moves in a circular path with a constant speed in the xy plane with the centre at the origin. When the object is at x = −2 m, its velocity is −(4 m/s)j^∧ . Then objects velocity at y = 2 m is (1) 4 m/s i^∧ (2) (−4 m/s) i^∧ Using this data, find objects acceleration when it is at y = 2 m (1) 8 m/s^2 i^∧ (2) −8 m/s^2 j^∧

$$\mathrm{An}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{is}\:\mathrm{at}\:{x}\:=\:−\mathrm{2}\:\mathrm{m},\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$$−\left(\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\overset{\wedge} {{j}}.\:\mathrm{Then}\:\mathrm{objects}\:\mathrm{velocity}\:\mathrm{at} \\ $$$${y}\:=\:\mathrm{2}\:\mathrm{m}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}\:\mathrm{m}/\mathrm{s}\:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{i}} \\ $$$$\mathrm{Using}\:\mathrm{this}\:\mathrm{data},\:\mathrm{find}\:\mathrm{objects}\:\mathrm{acceleration} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:{y}\:=\:\mathrm{2}\:\mathrm{m} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{j}} \\ $$

Question Number 16502 Answers: 1 Comments: 0

The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector. (True/False)

$$\mathrm{The}\:\mathrm{acceleration}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in} \\ $$$$\mathrm{uniform}\:\mathrm{circular}\:\mathrm{motion}\:\mathrm{averaged}\:\mathrm{over} \\ $$$$\mathrm{one}\:\mathrm{cycle}\:\mathrm{is}\:\mathrm{a}\:\mathrm{null}\:\mathrm{vector}.\:\left(\mathrm{True}/\mathrm{False}\right) \\ $$

Question Number 16501 Answers: 1 Comments: 0

3+3

$$\mathrm{3}+\mathrm{3} \\ $$

Question Number 16499 Answers: 1 Comments: 0

A particle is moving in parabolic path x^2 = y, with constant speed u. Find the acceleration of the particle when it crossess origin. Also find the radius of curvature at origin.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{parabolic}\:\mathrm{path} \\ $$$${x}^{\mathrm{2}} \:=\:{y},\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{u}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it} \\ $$$$\mathrm{crossess}\:\mathrm{origin}.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of} \\ $$$$\mathrm{curvature}\:\mathrm{at}\:\mathrm{origin}. \\ $$

Question Number 16493 Answers: 0 Comments: 1

Find L{tcosh2t}

$${Find}\:{L}\left\{{tcosh}\mathrm{2}{t}\right\} \\ $$

Question Number 16492 Answers: 1 Comments: 0

solve Mdc/dt=P+km−Cm where M.P.m and k are constant solve the equation given C=a when t=o

$${solve} \\ $$$${Mdc}/{dt}={P}+{km}−{Cm}\:{where}\:{M}.{P}.{m}\:{and} \\ $$$${k}\:{are}\:{constant} \\ $$$${solve}\:{the}\:{equation}\:{given}\:{C}={a}\:{when}\:{t}={o} \\ $$

Question Number 16491 Answers: 0 Comments: 0

Men are running in a line along a road with velocity 9 km/hr behind one another at equal distances of 20 m. Cyclists are also riding along the same line in the same direction at 18 km/hr at equal intervals of 30 m. The speed with which an observer must travel along the road in opposite direction of so that whenever he meets a runner he also meets a cyclist is (1) 9 km/h (2) 12 km/h (3) 18 km/h (4) 6 km/h

$$\mathrm{Men}\:\mathrm{are}\:\mathrm{running}\:\mathrm{in}\:\mathrm{a}\:\mathrm{line}\:\mathrm{along}\:\mathrm{a}\:\mathrm{road} \\ $$$$\mathrm{with}\:\mathrm{velocity}\:\mathrm{9}\:\mathrm{km}/\mathrm{hr}\:\mathrm{behind}\:\mathrm{one} \\ $$$$\mathrm{another}\:\mathrm{at}\:\mathrm{equal}\:\mathrm{distances}\:\mathrm{of}\:\mathrm{20}\:\mathrm{m}. \\ $$$$\mathrm{Cyclists}\:\mathrm{are}\:\mathrm{also}\:\mathrm{riding}\:\mathrm{along}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{line}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{at}\:\mathrm{18}\:\mathrm{km}/\mathrm{hr} \\ $$$$\mathrm{at}\:\mathrm{equal}\:\mathrm{intervals}\:\mathrm{of}\:\mathrm{30}\:\mathrm{m}.\:\mathrm{The}\:\mathrm{speed} \\ $$$$\mathrm{with}\:\mathrm{which}\:\mathrm{an}\:\mathrm{observer}\:\mathrm{must}\:\mathrm{travel} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{road}\:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}\:\mathrm{of} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{whenever}\:\mathrm{he}\:\mathrm{meets}\:\mathrm{a}\:\mathrm{runner}\:\mathrm{he} \\ $$$$\mathrm{also}\:\mathrm{meets}\:\mathrm{a}\:\mathrm{cyclist}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{9}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{12}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{18}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{6}\:\mathrm{km}/\mathrm{h} \\ $$

Question Number 16497 Answers: 1 Comments: 0

Two particles are revolving on two coplanar circles with constant angular velocities ω_1 and ω_2 respectively. Their time periods are T_1 and T_2 then prove that the time taken by second particle to complete one revolution more than the first particle, T, is given by T = ((T_1 T_2 )/(T_1 − T_2 ))

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{are}\:\mathrm{revolving}\:\mathrm{on}\:\mathrm{two} \\ $$$$\mathrm{coplanar}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular} \\ $$$$\mathrm{velocities}\:\omega_{\mathrm{1}} \:\mathrm{and}\:\omega_{\mathrm{2}} \:\mathrm{respectively}.\:\mathrm{Their} \\ $$$$\mathrm{time}\:\mathrm{periods}\:\mathrm{are}\:{T}_{\mathrm{1}} \:\mathrm{and}\:{T}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{second}\:\mathrm{particle} \\ $$$$\mathrm{to}\:\mathrm{complete}\:\mathrm{one}\:\mathrm{revolution}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{particle},\:{T},\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${T}\:=\:\frac{{T}_{\mathrm{1}} {T}_{\mathrm{2}} }{{T}_{\mathrm{1}} \:−\:{T}_{\mathrm{2}} } \\ $$

Question Number 16489 Answers: 1 Comments: 0

Rain is falling vertically with a speed of 4 m/s. After some time, wind starts blowing with a speed of 3 m/s in the north to south direction. In order to protect himself from rain, a man standing on the ground should hold his umbrella at an angle θ given by (1) θ = tan^(−1) 3/4 with the vertical towards south (2) θ = tan^(−1) 3/4 with the vertical towards north (3) θ = cot^(−1) 3/4 with the vertical towards south (1) θ = cot^(−1) 3/4 with the vertical towards north

$$\mathrm{Rain}\:\mathrm{is}\:\mathrm{falling}\:\mathrm{vertically}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{4}\:\mathrm{m}/\mathrm{s}.\:\mathrm{After}\:\mathrm{some}\:\mathrm{time},\:\mathrm{wind}\:\mathrm{starts} \\ $$$$\mathrm{blowing}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{3}\:\mathrm{m}/\mathrm{s}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{north}\:\mathrm{to}\:\mathrm{south}\:\mathrm{direction}.\:\mathrm{In}\:\mathrm{order}\:\mathrm{to} \\ $$$$\mathrm{protect}\:\mathrm{himself}\:\mathrm{from}\:\mathrm{rain},\:\mathrm{a}\:\mathrm{man} \\ $$$$\mathrm{standing}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{should}\:\mathrm{hold}\:\mathrm{his} \\ $$$$\mathrm{umbrella}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{given}\:\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{south} \\ $$$$\left(\mathrm{2}\right)\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{north} \\ $$$$\left(\mathrm{3}\right)\:\theta\:=\:\mathrm{cot}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{south} \\ $$$$\left(\mathrm{1}\right)\:\theta\:=\:\mathrm{cot}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{towards}\:\mathrm{north} \\ $$

Question Number 16485 Answers: 1 Comments: 0

If tan^2 θ=2 tan^2 φ+1, then cos 2θ+sin^2 φ equals

$$\mathrm{If}\:\mathrm{tan}^{\mathrm{2}} \theta=\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \phi+\mathrm{1},\:\mathrm{then}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{sin}^{\mathrm{2}} \phi \\ $$$$\mathrm{equals} \\ $$

Question Number 16483 Answers: 0 Comments: 12

Question Number 16475 Answers: 1 Comments: 0

use L{f′(t)} to find laplace transform of d^2 x/dt^2 +n^2 x=kcoswt given that x=0 and dx/dt=0 when t=0 PLZ HELP.

$${use}\:{L}\left\{{f}'\left({t}\right)\right\}\:{to}\:{find}\:{laplace}\: \\ $$$${transform}\:{of}\:{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}={kcoswt} \\ $$$${given}\:{that}\:{x}=\mathrm{0}\:{and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0} \\ $$$${PLZ}\:{HELP}. \\ $$$$ \\ $$

Question Number 16466 Answers: 1 Comments: 0

Question Number 16464 Answers: 0 Comments: 2

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