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Question Number 24539 Answers: 0 Comments: 1

Question Number 24526 Answers: 2 Comments: 1

Question Number 24524 Answers: 1 Comments: 0

Question Number 24520 Answers: 2 Comments: 0

A particle moves in a straight line along x-axis. At t = 0 it passes origin with some velocity towards positive x-axis and with an acceleration a which is given as, a = − Kx, where x is in metre and K is a positive constant. The time at which its velocity becomes half of its value at t = 0 for the first time, is

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{along} \\ $$$${x}-\mathrm{axis}.\:\mathrm{At}\:{t}\:=\:\mathrm{0}\:\mathrm{it}\:\mathrm{passes}\:\mathrm{origin}\:\mathrm{with} \\ $$$$\mathrm{some}\:\mathrm{velocity}\:\mathrm{towards}\:\mathrm{positive}\:{x}-\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}\:{a}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{as},\:{a}\:=\:−\:{Kx},\:\mathrm{where}\:{x}\:\mathrm{is}\:\mathrm{in}\:\mathrm{metre} \\ $$$$\mathrm{and}\:{K}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}.\:\mathrm{The}\:\mathrm{time} \\ $$$$\mathrm{at}\:\mathrm{which}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{becomes}\:\mathrm{half}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{value}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{time},\:\mathrm{is} \\ $$

Question Number 24490 Answers: 0 Comments: 2

I hold my hand behind my back, and secretly hold up n fingers (0≤n≤5). I then as you, am I holding up k fingers? Where k is also a random number 0≤k≤5. You randomly guess Yes or No. What is the probability you guess correctly?

$$\mathrm{I}\:\mathrm{hold}\:\mathrm{my}\:\mathrm{hand}\:\mathrm{behind}\:\mathrm{my}\:\mathrm{back},\:\mathrm{and}\:\mathrm{secretly} \\ $$$$\mathrm{hold}\:\mathrm{up}\:{n}\:\mathrm{fingers}\:\left(\mathrm{0}\leqslant{n}\leqslant\mathrm{5}\right). \\ $$$$\: \\ $$$$\mathrm{I}\:\mathrm{then}\:\mathrm{as}\:\mathrm{you},\:\mathrm{am}\:\mathrm{I}\:\mathrm{holding}\:\mathrm{up}\:{k}\:\mathrm{fingers}? \\ $$$$\mathrm{Where}\:{k}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{random}\:\mathrm{number}\:\mathrm{0}\leqslant{k}\leqslant\mathrm{5}. \\ $$$$\: \\ $$$$\mathrm{You}\:\mathrm{randomly}\:\mathrm{guess}\:\mathrm{Yes}\:\mathrm{or}\:\mathrm{No}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{you}\:\mathrm{guess}\:\mathrm{correctly}? \\ $$

Question Number 24482 Answers: 0 Comments: 4

Neglecting friction and mass of pulleys, what is the acceleration of mass B?

$$\mathrm{Neglecting}\:\mathrm{friction}\:\mathrm{and}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{pulleys}, \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{mass}\:{B}? \\ $$

Question Number 24479 Answers: 0 Comments: 0

Describe the energy transformations that take place when a skier starts sking down a hill, but after a time is brought to rest by striking a snowdrift.

$$\mathrm{Describe}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{transformations} \\ $$$$\mathrm{that}\:\mathrm{take}\:\mathrm{place}\:\mathrm{when}\:\mathrm{a}\:\mathrm{skier}\:\mathrm{starts} \\ $$$$\mathrm{sking}\:\mathrm{down}\:\mathrm{a}\:\mathrm{hill},\:\mathrm{but}\:\mathrm{after}\:\mathrm{a}\:\mathrm{time}\:\mathrm{is} \\ $$$$\mathrm{brought}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{by}\:\mathrm{striking}\:\mathrm{a}\:\mathrm{snowdrift}. \\ $$

Question Number 24477 Answers: 1 Comments: 2

A particle moving horizonatally collides perpendiculrly at one end of a rod having equal mass and placed on a horizontal surface. The book says that particle will continue to move along the same direction regardless of value of e (coefficient of restitution). I did not understand the logic. please help.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{horizonatally} \\ $$$$\mathrm{collides}\:\mathrm{perpendiculrly}\:\mathrm{at}\:\mathrm{one}\:\mathrm{end} \\ $$$$\mathrm{of}\:\:\mathrm{a}\:\mathrm{rod}\:\mathrm{having}\:\mathrm{equal}\:\mathrm{mass}\:\mathrm{and} \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{surface}. \\ $$$$\mathrm{The}\:\mathrm{book}\:\mathrm{says}\:\mathrm{that}\:\mathrm{particle}\:\mathrm{will} \\ $$$$\mathrm{continue}\:\mathrm{to}\:\mathrm{move}\:\mathrm{along}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{direction}\:\mathrm{regardless}\:\mathrm{of}\:\mathrm{value}\:\mathrm{of} \\ $$$${e}\:\left(\mathrm{coefficient}\:\mathrm{of}\:\mathrm{restitution}\right). \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{logic}. \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$

Question Number 24469 Answers: 0 Comments: 2

Let 2x + 3y + 4z = 9, x, y, z > 0 then the maximum value of (1 + x)^2 (2 + y)^3 (4 + z)^4 is

$$\mathrm{Let}\:\mathrm{2}{x}\:+\:\mathrm{3}{y}\:+\:\mathrm{4}{z}\:=\:\mathrm{9},\:{x},\:{y},\:{z}\:>\:\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}\:+\:{x}\right)^{\mathrm{2}} \:\left(\mathrm{2}\:+\:{y}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{4}\:+\:{z}\right)^{\mathrm{4}} \:\mathrm{is} \\ $$

Question Number 24465 Answers: 0 Comments: 3

Two blocks of masses m_1 and m_2 are placed in contact with each other on a horizontal platform. The coefficient of friction between the platform and the two blocks is the same. The platform moves with an acceleration. The force of interaction between the blocks is

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:{m}_{\mathrm{1}} \:\mathrm{and}\:{m}_{\mathrm{2}} \:\mathrm{are} \\ $$$$\mathrm{placed}\:\mathrm{in}\:\mathrm{contact}\:\mathrm{with}\:\mathrm{each}\:\mathrm{other}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{platform}.\:\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of} \\ $$$$\mathrm{friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{platform}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{two}\:\mathrm{blocks}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}.\:\mathrm{The}\:\mathrm{platform} \\ $$$$\mathrm{moves}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}.\:\mathrm{The}\:\mathrm{force} \\ $$$$\mathrm{of}\:\mathrm{interaction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{blocks}\:\mathrm{is} \\ $$

Question Number 24460 Answers: 0 Comments: 3

A uniform rod of AB of mass m=1.12 kg and lengtj l=100cm is placed on a sharp support O such that AO=a=40cm. To keep the rod horizontal, its end A is ties with a thread. Calculate reaction of support O on the rod when the thread is burnt. (g=10 m∙s^(−2) )

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{rod}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{of}\:\mathrm{mass} \\ $$$$\mathrm{m}=\mathrm{1}.\mathrm{12}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{lengtj}\:\mathrm{l}=\mathrm{100cm}\:\mathrm{is} \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{sharp}\:\mathrm{support}\:\mathrm{O}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{AO}={a}=\mathrm{40}{cm}.\:\mathrm{To}\:\mathrm{keep}\:\mathrm{the}\:\mathrm{rod} \\ $$$$\mathrm{horizontal},\:\mathrm{its}\:\mathrm{end}\:\mathrm{A}\:\mathrm{is}\:\mathrm{ties}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{thread}.\:\mathrm{Calculate}\:\mathrm{reaction}\:\mathrm{of}\:\mathrm{support} \\ $$$$\mathrm{O}\:\mathrm{on}\:\mathrm{the}\:\mathrm{rod}\:\mathrm{when}\:\mathrm{the}\:\mathrm{thread}\:\mathrm{is}\:\mathrm{burnt}. \\ $$$$\left({g}=\mathrm{10}\:\mathrm{m}\centerdot\mathrm{s}^{−\mathrm{2}} \right) \\ $$

Question Number 24459 Answers: 2 Comments: 0

If the sides of a rectangle are increased each by 10%, find the % increase in its diagonal.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{are}\:\mathrm{increased} \\ $$$$\mathrm{each}\:\mathrm{by}\:\mathrm{10\%},\:\mathrm{find}\:\mathrm{the}\:\%\:\mathrm{increase}\:\mathrm{in}\:\mathrm{its} \\ $$$$\mathrm{diagonal}. \\ $$

Question Number 24456 Answers: 1 Comments: 0

The product of a monomial by a binomial is a

$${The}\:{product}\:{of}\:{a}\:{monomial}\:{by}\:{a}\:{binomial}\:{is}\:{a} \\ $$

Question Number 24434 Answers: 0 Comments: 0

Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R + δ). They are now brought in contact (δ → 0). (a) Show the frictional forces just after contact. (b) Identify forces and torques external to the system just after contact. (c) What would be the ratio of final angular velocities when friction ceases?

$$\mathrm{Two}\:\mathrm{cylindrical}\:\mathrm{hollow}\:\mathrm{drums}\:\mathrm{of}\:\mathrm{radii} \\ $$$${R}\:\mathrm{and}\:\mathrm{2}{R},\:\mathrm{and}\:\mathrm{of}\:\mathrm{a}\:\mathrm{common}\:\mathrm{height}\:{h}, \\ $$$$\mathrm{are}\:\mathrm{rotating}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{velocities}\:\omega \\ $$$$\left(\mathrm{anti}-\mathrm{clockwise}\right)\:\mathrm{and}\:\omega\:\left(\mathrm{clockwise}\right), \\ $$$$\mathrm{respectively}.\:\mathrm{Their}\:\mathrm{axes},\:\mathrm{fixed}\:\mathrm{are} \\ $$$$\mathrm{parallel}\:\mathrm{and}\:\mathrm{in}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{plane} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\left(\mathrm{3}{R}\:+\:\delta\right).\:\mathrm{They}\:\mathrm{are}\:\mathrm{now} \\ $$$$\mathrm{brought}\:\mathrm{in}\:\mathrm{contact}\:\left(\delta\:\rightarrow\:\mathrm{0}\right). \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{the}\:\mathrm{frictional}\:\mathrm{forces}\:\mathrm{just}\:\mathrm{after} \\ $$$$\mathrm{contact}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Identify}\:\mathrm{forces}\:\mathrm{and}\:\mathrm{torques}\:\mathrm{external} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{system}\:\mathrm{just}\:\mathrm{after}\:\mathrm{contact}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{What}\:\mathrm{would}\:\mathrm{be}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{final} \\ $$$$\mathrm{angular}\:\mathrm{velocities}\:\mathrm{when}\:\mathrm{friction}\:\mathrm{ceases}? \\ $$

Question Number 24436 Answers: 0 Comments: 0

For a reversible reaction A ⇌ B. Find K_(eq) at 2727°C temperature. Given : Δ_r H° = −30 kJ mol^(−1) (at 2727°C) Δ_r S° = 10 JK^(−1) (at 2727°C) R = 8.314 JK^(−1) mol^(−1)

$$\mathrm{For}\:\mathrm{a}\:\mathrm{reversible}\:\mathrm{reaction}\:\mathrm{A}\:\rightleftharpoons\:\mathrm{B}.\:\mathrm{Find} \\ $$$$\mathrm{K}_{\mathrm{eq}} \:\mathrm{at}\:\mathrm{2727}°\mathrm{C}\:\mathrm{temperature}. \\ $$$$\mathrm{Given}\::\:\Delta_{\mathrm{r}} \mathrm{H}°\:=\:−\mathrm{30}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \:\left(\mathrm{at}\:\mathrm{2727}°\mathrm{C}\right) \\ $$$$\Delta_{\mathrm{r}} \mathrm{S}°\:=\:\mathrm{10}\:\mathrm{JK}^{−\mathrm{1}} \:\left(\mathrm{at}\:\mathrm{2727}°\mathrm{C}\right) \\ $$$$\mathrm{R}\:=\:\mathrm{8}.\mathrm{314}\:\mathrm{JK}^{−\mathrm{1}} \:\mathrm{mol}^{−\mathrm{1}} \\ $$

Question Number 24428 Answers: 1 Comments: 0

In a race on a track of a certain length, A beats B by 20 m. When A was 10 m ahead of the mid−point of the track B was 2 m behind it. Find the length of the track (in m).

$$\mathrm{In}\:\mathrm{a}\:\mathrm{race}\:\mathrm{on}\:\mathrm{a}\:\mathrm{track}\:\:\mathrm{of}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{length}, \\ $$$$\mathrm{A}\:\:\mathrm{beats}\:\mathrm{B}\:\mathrm{by}\:\mathrm{20}\:\mathrm{m}.\:\mathrm{When}\:\mathrm{A}\:\mathrm{was}\:\mathrm{10}\:\mathrm{m} \\ $$$$\mathrm{ahead}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{track}\:\mathrm{B} \\ $$$$\mathrm{was}\:\mathrm{2}\:\mathrm{m}\:\mathrm{behind}\:\mathrm{it}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{track}\:\left(\mathrm{in}\:\mathrm{m}\right). \\ $$

Question Number 24426 Answers: 1 Comments: 0

A number is multiplied by 2(1/3) times itself and then 61 is subtracted from the product obtained. If the final result is 9200, then the number is

$$\mathrm{A}\:\mathrm{number}\:\mathrm{is}\:\mathrm{multiplied}\:\mathrm{by}\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{times} \\ $$$$\mathrm{itself}\:\mathrm{and}\:\mathrm{then}\:\mathrm{61}\:\mathrm{is}\:\mathrm{subtracted}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{obtained}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{final}\:\mathrm{result} \\ $$$$\mathrm{is}\:\mathrm{9200},\:\mathrm{then}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is} \\ $$

Question Number 24447 Answers: 0 Comments: 8

Figure shows two discs of same mass m. They are rigidly attached to a spring of stiffness k. The system is in equilibrium. From this equilibrium position, the upper disc is pressed down slowly by a distance x and released. Find the minimum value of x, if the lower disc is just lifted off the ground.

$${Figure}\:{shows}\:{two}\:{discs}\:{of}\:{same}\:{mass} \\ $$$${m}.\:{They}\:{are}\:{rigidly}\:{attached}\:{to}\:{a}\:{spring} \\ $$$${of}\:{stiffness}\:{k}.\:{The}\:{system}\:{is}\:{in} \\ $$$${equilibrium}.\:{From}\:{this}\:{equilibrium} \\ $$$${position},\:{the}\:{upper}\:{disc}\:{is}\:{pressed} \\ $$$${down}\:{slowly}\:{by}\:{a}\:{distance}\:{x}\:{and} \\ $$$${released}.\:{Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${x},\:{if}\:{the}\:{lower}\:{disc}\:{is}\:{just}\:{lifted}\:{off} \\ $$$${the}\:{ground}. \\ $$

Question Number 24443 Answers: 2 Comments: 0

Solve for x: (10^(−4) x)^x =4×10^(−8)

$${Solve}\:{for}\:{x}: \\ $$$$\left(\mathrm{10}^{−\mathrm{4}} {x}\right)^{{x}} =\mathrm{4}×\mathrm{10}^{−\mathrm{8}} \\ $$

Question Number 24417 Answers: 1 Comments: 0

The time taken by a train l metres long running at x km/h to pass a man who is running at y km/h in the direction opposite to that of the train = The time taken to cover l metres at _____ km/h.

$$\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{a}\:\mathrm{train}\:{l}\:\mathrm{metres}\:\mathrm{long} \\ $$$$\mathrm{running}\:\mathrm{at}\:{x}\:\mathrm{km}/\mathrm{h}\:\mathrm{to}\:\mathrm{pass}\:\mathrm{a}\:\mathrm{man}\:\mathrm{who} \\ $$$$\mathrm{is}\:\mathrm{running}\:\mathrm{at}\:{y}\:\mathrm{km}/\mathrm{h}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{train}\:=\:\mathrm{The}\:\mathrm{time} \\ $$$$\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:{l}\:\mathrm{metres}\:\mathrm{at}\:\_\_\_\_\_\:\mathrm{km}/\mathrm{h}. \\ $$

Question Number 24438 Answers: 0 Comments: 2

Find the sum to infinite terms of the series (x/(1−x^2 ))+(x^2 /(1−x^4 ))+(x^4 /(1−x^8 ))+......

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinite}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{series}\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{8}} }+...... \\ $$

Question Number 24411 Answers: 1 Comments: 0

Question Number 24405 Answers: 0 Comments: 0

Find the horizontal assymptot lim_(x→∞ ) (((3x^2 +4))^(1/6) /((1−2x^3 ))^(1/9) )

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{assymptot} \\ $$$$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{\mathrm{3x}^{\mathrm{2}} +\mathrm{4}}}{\sqrt[{\mathrm{9}}]{\mathrm{1}−\mathrm{2x}^{\mathrm{3}} }} \\ $$$$ \\ $$

Question Number 24404 Answers: 0 Comments: 0

put a_n = (1/(2n))∙(3/(2n))∙(5/(2n))∙ ... ∙((2n−1)/(2n))∙e^n lim_(n→∞) a_n = ?

$${put}\:{a}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{3}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{5}}{\mathrm{2}{n}}\centerdot\:...\:\centerdot\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\centerdot{e}^{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\:? \\ $$$$ \\ $$

Question Number 24387 Answers: 1 Comments: 0

Two particle move along an x−axis. The position of particle 1 is given; by x=6.00t^2 +3.00t+2.00((m/s)) the acceleration of particle 2 is given by a=−8.00t((m/s^2 )) and,at t=0,its velocity is 20 ((m/s)).when the velocities of the particles match, what is their velocity? plzz help

Question Number 24379 Answers: 2 Comments: 0

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