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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

Question Number 17493 Answers: 0 Comments: 0

What is logarithm series?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{logarithm}\:\mathrm{series}? \\ $$

Question Number 16457 Answers: 1 Comments: 0

Find the range of f(x) = (3/(2 − x^2 ))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}\:−\:{x}^{\mathrm{2}} } \\ $$

Question Number 16455 Answers: 2 Comments: 0

The speed of a projectile when it is at its greatest height is (√(2/5)) times its speed at half the maximum height. What is its angle of projection?

$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{its}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{is}\:\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{times}\:\mathrm{its} \\ $$$$\mathrm{speed}\:\mathrm{at}\:\mathrm{half}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{its}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}? \\ $$

Question Number 16451 Answers: 0 Comments: 0

The sixth term of an AP is 2, and its common difference is greater than one. The value of the common difference of the progression so that the product of the first, fourth and fifth terms is greatest is

$$\mathrm{The}\:\mathrm{sixth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{2},\:\mathrm{and}\:\mathrm{its} \\ $$$$\mathrm{common}\:\mathrm{difference}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{one}. \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{progression}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first},\:\mathrm{fourth}\:\mathrm{and}\:\mathrm{fifth}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{greatest}\:\mathrm{is} \\ $$

Question Number 16430 Answers: 1 Comments: 2

In ΔABC with usual notation (r_1 /(bc)) + (r_2 /(ca)) + (r_3 /(ab)) is (1) (1/r) − (1/R) (2) (1/r) − (1/(2R)) (3) (1/r) + (1/(2R)) (4) (1/r) + (1/R)

$$\mathrm{In}\:\Delta{ABC}\:\mathrm{with}\:\mathrm{usual}\:\mathrm{notation} \\ $$$$\frac{{r}_{\mathrm{1}} }{{bc}}\:+\:\frac{{r}_{\mathrm{2}} }{{ca}}\:+\:\frac{{r}_{\mathrm{3}} }{{ab}}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{{R}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{{R}} \\ $$

Question Number 17647 Answers: 0 Comments: 5

ABC is a triangular park with AB = AC = 100 m. A clock tower is situated at the midpoint of BC. The angles of elevation of top of the tower at A and B are cot^(−1) (3.2) and cosec^(−1) (2.6) respectively. The height of tower is

$${ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangular}\:\mathrm{park}\:\mathrm{with}\:{AB}\:= \\ $$$${AC}\:=\:\mathrm{100}\:\mathrm{m}.\:\mathrm{A}\:\mathrm{clock}\:\mathrm{tower}\:\mathrm{is}\:\mathrm{situated} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{BC}.\:\mathrm{The}\:\mathrm{angles}\:\mathrm{of} \\ $$$$\mathrm{elevation}\:\mathrm{of}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower}\:\mathrm{at}\:{A}\:\mathrm{and} \\ $$$${B}\:\mathrm{are}\:\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{3}.\mathrm{2}\right)\:\mathrm{and}\:\mathrm{cosec}^{−\mathrm{1}} \left(\mathrm{2}.\mathrm{6}\right) \\ $$$$\mathrm{respectively}.\:\mathrm{The}\:\mathrm{height}\:\mathrm{of}\:\mathrm{tower}\:\mathrm{is} \\ $$

Question Number 16419 Answers: 1 Comments: 0

3 cubes of metal whose edges are 3,4 and 5 respectively are melted and formed into a single cube. If there be no loss of metal in the process find the side of the new cube.

$$\mathrm{3}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{whose}\:\mathrm{edges}\:\mathrm{are}\:\mathrm{3},\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{5}\:\mathrm{respectively}\:\mathrm{are}\:\mathrm{melted}\:\mathrm{and} \\ $$$$\mathrm{formed}\:\mathrm{into}\:\mathrm{a}\:\mathrm{single}\:\mathrm{cube}.\:\mathrm{If}\:\mathrm{there}\:\mathrm{be} \\ $$$$\mathrm{no}\:\mathrm{loss}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{process}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{new}\:\mathrm{cube}. \\ $$

Question Number 16418 Answers: 1 Comments: 0

A solid sphere of radius 4 cm is melted and recast into ′n′ solid hemispheres of radius 2 cm each. Find n.

$$\mathrm{A}\:\mathrm{solid}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{4}\:\mathrm{cm}\:\mathrm{is}\:\mathrm{melted} \\ $$$$\mathrm{and}\:\mathrm{recast}\:\mathrm{into}\:'{n}'\:\mathrm{solid}\:\mathrm{hemispheres}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{2}\:\mathrm{cm}\:\mathrm{each}.\:\mathrm{Find}\:{n}. \\ $$

Question Number 16409 Answers: 3 Comments: 9

Question Number 16406 Answers: 0 Comments: 0

D={(x,y)∣∣x∣+∣y∣≤1,∣x∣+∣y∣≥0.5} ∫∫_(D) ln (x^2 +y^2 )dxdy a)≥0; b)≤0; c)=0; d)non-existent.

$$\mathrm{D}=\left\{\left({x},\mathrm{y}\right)\mid\mid{x}\mid+\mid\mathrm{y}\mid\leqslant\mathrm{1},\mid{x}\mid+\mid{y}\mid\geqslant\mathrm{0}.\mathrm{5}\right\} \\ $$$$\underset{\mathrm{D}} {\int\int}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy} \\ $$$$\left.\mathrm{a}\right)\geqslant\mathrm{0}; \\ $$$$\left.\mathrm{b}\right)\leqslant\mathrm{0}; \\ $$$$\left.\mathrm{c}\right)=\mathrm{0}; \\ $$$$\left.\mathrm{d}\right)\mathrm{non}-\mathrm{existent}. \\ $$

Question Number 16401 Answers: 1 Comments: 0

v=2i+2j+5k r=i+9j−8k Find 𝛚 I can do ((r×v)/r^2 )=𝛚 and i get 𝛚= ((61i−21j−16k)/(146)) but i dont get w×r=v. why?

$$\boldsymbol{{v}}=\mathrm{2}\boldsymbol{{i}}+\mathrm{2}\boldsymbol{{j}}+\mathrm{5}\boldsymbol{{k}} \\ $$$$\boldsymbol{{r}}=\boldsymbol{{i}}+\mathrm{9}\boldsymbol{{j}}−\mathrm{8}\boldsymbol{{k}} \\ $$$$\mathrm{Find}\:\boldsymbol{\omega} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{do}\:\frac{\boldsymbol{{r}}×\boldsymbol{{v}}}{{r}^{\mathrm{2}} }=\boldsymbol{\omega} \\ $$$$\mathrm{and}\:\mathrm{i}\:\mathrm{get}\:\boldsymbol{\omega}=\:\frac{\mathrm{61}\boldsymbol{{i}}−\mathrm{21}\boldsymbol{{j}}−\mathrm{16}\boldsymbol{{k}}}{\mathrm{146}} \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{get}\:\boldsymbol{{w}}×\boldsymbol{{r}}=\boldsymbol{{v}}. \\ $$$${why}? \\ $$

Question Number 16392 Answers: 3 Comments: 0

Question Number 16387 Answers: 0 Comments: 0

Why sinx is a power series?

$${Why}\:{sinx}\:{is}\:{a}\:{power}\:{series}? \\ $$

Question Number 16383 Answers: 2 Comments: 0

If P : Q : R = 2 : 3 : 4 and P^2 +Q^2 +R^2 =11600, then find (P+Q−R).

$$\mathrm{If}\:\mathrm{P}\::\:\mathrm{Q}\::\:\mathrm{R}\:=\:\mathrm{2}\::\:\mathrm{3}\::\:\mathrm{4}\:\mathrm{and}\:\mathrm{P}^{\mathrm{2}} +\mathrm{Q}^{\mathrm{2}} +\mathrm{R}^{\mathrm{2}} =\mathrm{11600}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\left(\mathrm{P}+\mathrm{Q}−\mathrm{R}\right). \\ $$

Question Number 16373 Answers: 1 Comments: 1

if Σ_(k=0) ^(200) i^k +Π_(p=1) ^(50) i^p =x+iy then..(x,y)is... a. (0,1) b. (1,−1) c. (2,3) d. (4,8)

$${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}... \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\ $$

Question Number 16364 Answers: 1 Comments: 0

Question Number 16363 Answers: 1 Comments: 3

A car drives due north at 50 km/hr. Wind blows due North-West at 50(√2) km/hr. In what direction, a flag hoisted on the roof of the car points?

$$\mathrm{A}\:\mathrm{car}\:\mathrm{drives}\:\mathrm{due}\:\mathrm{north}\:\mathrm{at}\:\mathrm{50}\:\mathrm{km}/\mathrm{hr}. \\ $$$$\mathrm{Wind}\:\mathrm{blows}\:\mathrm{due}\:\mathrm{North}-\mathrm{West}\:\mathrm{at}\:\mathrm{50}\sqrt{\mathrm{2}} \\ $$$$\mathrm{km}/\mathrm{hr}.\:\mathrm{In}\:\mathrm{what}\:\mathrm{direction},\:\mathrm{a}\:\mathrm{flag} \\ $$$$\mathrm{hoisted}\:\mathrm{on}\:\mathrm{the}\:\mathrm{roof}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}\:\mathrm{points}? \\ $$

Question Number 16360 Answers: 1 Comments: 0

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