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Question Number 22304    Answers: 1   Comments: 0

A particle moves in a straight line along x-axis. At t = 0, it is released from rest at x = a. Acceleration of the particle varies as (d^2 x/dt^2 ) = −(k/x^2 ) , where k is a positive constant. Time required by particle to reach the origin will be

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{along} \\ $$$${x}-\mathrm{axis}.\:\mathrm{At}\:{t}\:=\:\mathrm{0},\:\mathrm{it}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from} \\ $$$$\mathrm{rest}\:\mathrm{at}\:{x}\:=\:{a}.\:\mathrm{Acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{varies}\:\mathrm{as}\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:=\:−\frac{{k}}{{x}^{\mathrm{2}} }\:,\:\mathrm{where}\:{k} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}. \\ $$$$\mathrm{Time}\:\mathrm{required}\:\mathrm{by}\:\mathrm{particle}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the} \\ $$$$\mathrm{origin}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 22302    Answers: 1   Comments: 0

A small bead of mass m is given an initial velocity of magnitude v_0 on a horizontal circular wire. If the coefficient of kinetic friction is μ_k , the determine the distance travelled before the collar comes to rest. (Given that radius of circular wire is R).

$$\mathrm{A}\:\mathrm{small}\:\mathrm{bead}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{given}\:\mathrm{an} \\ $$$$\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{magnitude}\:{v}_{\mathrm{0}} \:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{circular}\:\mathrm{wire}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{kinetic}\:\mathrm{friction}\:\mathrm{is}\:\mu_{\mathrm{k}} ,\:\mathrm{the} \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{before} \\ $$$$\mathrm{the}\:\mathrm{collar}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}.\:\left(\mathrm{Given}\:\mathrm{that}\right. \\ $$$$\left.\mathrm{radius}\:\mathrm{of}\:\mathrm{circular}\:\mathrm{wire}\:\mathrm{is}\:{R}\right). \\ $$

Question Number 22286    Answers: 1   Comments: 5

A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire B is (a) always radially outwards (b) always radially inwards (c) radially outwards initially and radially inwards later (d) radially inwards initially and radially outwards later.

$$\mathrm{A}\:\mathrm{wire},\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{hole} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{small}\:\mathrm{bead},\:\mathrm{is}\:\mathrm{bent}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of} \\ $$$$\mathrm{quarter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}.\:\mathrm{The}\:\mathrm{wire}\:\mathrm{is}\:\mathrm{fixed} \\ $$$$\mathrm{vertically}\:\mathrm{on}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{bead}\:\mathrm{is} \\ $$$$\mathrm{released}\:\mathrm{from}\:\mathrm{near}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire} \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{slides}\:\mathrm{along}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{without} \\ $$$$\mathrm{friction}.\:\mathrm{As}\:\mathrm{the}\:\mathrm{bead}\:\mathrm{moves}\:\mathrm{from}\:{A}\:\mathrm{to}\:{B}, \\ $$$$\mathrm{the}\:\mathrm{force}\:\mathrm{it}\:\mathrm{applies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{wire}\:{B}\:\mathrm{is} \\ $$$$\left({a}\right)\:\mathrm{always}\:\mathrm{radially}\:\mathrm{outwards} \\ $$$$\left({b}\right)\:\mathrm{always}\:\mathrm{radially}\:\mathrm{inwards} \\ $$$$\left({c}\right)\:\mathrm{radially}\:\mathrm{outwards}\:\mathrm{initially}\:\mathrm{and} \\ $$$$\mathrm{radially}\:\mathrm{inwards}\:\mathrm{later} \\ $$$$\left({d}\right)\:\mathrm{radially}\:\mathrm{inwards}\:\mathrm{initially}\:\mathrm{and} \\ $$$$\mathrm{radially}\:\mathrm{outwards}\:\mathrm{later}. \\ $$

Question Number 22279    Answers: 0   Comments: 0

Question Number 22299    Answers: 0   Comments: 0

If a certain mass of gas is made to undergo separately adiabatic and isothermal expansions to the same pressure, starting from the same initial conditions of temperature and pressure, then as compared to that of isothermal expansion, in the case of adiabatic expansion, the final (1) Volume and temperature will be higher (2) Volume and temperature will be lower (3) Temperature will be lower but the final volume will be higher (4) Volume will be lower but the final temperature will be higher

$$\mathrm{If}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{gas}\:\mathrm{is}\:\mathrm{made}\:\mathrm{to} \\ $$$$\mathrm{undergo}\:\mathrm{separately}\:\mathrm{adiabatic}\:\mathrm{and} \\ $$$$\mathrm{isothermal}\:\mathrm{expansions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{pressure},\:\mathrm{starting}\:\mathrm{from}\:\mathrm{the}\:\mathrm{same}\:\mathrm{initial} \\ $$$$\mathrm{conditions}\:\mathrm{of}\:\mathrm{temperature}\:\mathrm{and}\:\mathrm{pressure}, \\ $$$$\mathrm{then}\:\mathrm{as}\:\mathrm{compared}\:\mathrm{to}\:\mathrm{that}\:\mathrm{of}\:\mathrm{isothermal} \\ $$$$\mathrm{expansion},\:\mathrm{in}\:\mathrm{the}\:\mathrm{case}\:\mathrm{of}\:\mathrm{adiabatic} \\ $$$$\mathrm{expansion},\:\mathrm{the}\:\mathrm{final} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Volume}\:\mathrm{and}\:\mathrm{temperature}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{higher} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Volume}\:\mathrm{and}\:\mathrm{temperature}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{lower} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Temperature}\:\mathrm{will}\:\mathrm{be}\:\mathrm{lower}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{final}\:\mathrm{volume}\:\mathrm{will}\:\mathrm{be}\:\mathrm{higher} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Volume}\:\mathrm{will}\:\mathrm{be}\:\mathrm{lower}\:\mathrm{but}\:\mathrm{the}\:\mathrm{final} \\ $$$$\mathrm{temperature}\:\mathrm{will}\:\mathrm{be}\:\mathrm{higher} \\ $$

Question Number 22273    Answers: 0   Comments: 1

Ant rided 30 meters in 12 seconds. How fast are rided ant?

$${Ant}\:{rided}\:\mathrm{30}\:{meters}\:{in}\:\mathrm{12}\:{seconds}.\:{How}\:{fast}\:{are}\:{rided}\:{ant}? \\ $$

Question Number 23106    Answers: 0   Comments: 0

Is 2(x+1) had a x=3

$${Is}\:\mathrm{2}\left({x}+\mathrm{1}\right)\:{had}\:{a}\:{x}=\mathrm{3} \\ $$

Question Number 22269    Answers: 1   Comments: 0

Product costs 599.99 zloty. Person have 433.94 zloty. How muh zloty is a rest or missing?

$${Product}\:{costs}\:\mathrm{599}.\mathrm{99}\:{zloty}.\:{Person}\:{have}\:\mathrm{433}.\mathrm{94}\:{zloty}. \\ $$$${How}\:{muh}\:{zloty}\:{is}\:{a}\:{rest}\:{or}\:{missing}? \\ $$

Question Number 22268    Answers: 1   Comments: 2

if tanx = (1/3) or other (except standard values) how to find x

$$\mathrm{if}\:\mathrm{tanx}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{or}\:\mathrm{other}\:\left(\mathrm{except}\:\mathrm{standard}\:\right. \\ $$$$\left.\mathrm{values}\right)\:\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{x} \\ $$

Question Number 22264    Answers: 1   Comments: 0

What is 1+5(2/3)

$${What}\:{is}\:\mathrm{1}+\mathrm{5}\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Question Number 22263    Answers: 2   Comments: 0

Question Number 22261    Answers: 1   Comments: 0

Slove ∫xtanx dx

$$\mathrm{Slove} \\ $$$$\int\mathrm{xtanx}\:\mathrm{dx} \\ $$

Question Number 22260    Answers: 2   Comments: 0

8^(x−1) +(3/4)x=1 solve for x Any idea?

$$\mathrm{8}^{\boldsymbol{{x}}−\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{4}}\boldsymbol{{x}}=\mathrm{1}\:\:\:\:\:\:\boldsymbol{{solve}}\:\boldsymbol{{for}}\:\boldsymbol{{x}} \\ $$$$\boldsymbol{{Any}}\:\boldsymbol{{idea}}? \\ $$

Question Number 22726    Answers: 1   Comments: 2

Question Number 22728    Answers: 0   Comments: 0

Question Number 22730    Answers: 1   Comments: 2

Question Number 22729    Answers: 2   Comments: 5

Question Number 22244    Answers: 1   Comments: 0

Solve the inequality : −9((x)^(1/4) )+(√x)+18 ≥ 0 .

$${Solve}\:{the}\:{inequality}\:: \\ $$$$\:−\mathrm{9}\left(\sqrt[{\mathrm{4}}]{{x}}\right)+\sqrt{{x}}+\mathrm{18}\:\geqslant\:\mathrm{0}\:. \\ $$

Question Number 22232    Answers: 1   Comments: 0

find (dy/dx) where sin^(−1) ((x/y))=x+y

$$\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{where}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{x}+\mathrm{y} \\ $$

Question Number 22731    Answers: 0   Comments: 0

Question Number 22247    Answers: 1   Comments: 0

I=∫(√)x^2 +a^2 dx

$$\mathrm{I}=\int\sqrt{}\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \:\mathrm{dx} \\ $$

Question Number 22220    Answers: 1   Comments: 0

The number of integral solutions of the equation 4log_(x/2) ((√x))+2log_(4x) (x^2 )= 3log_(2x) (x^3 ) is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4log}_{{x}/\mathrm{2}} \left(\sqrt{{x}}\right)+\mathrm{2log}_{\mathrm{4}{x}} \left({x}^{\mathrm{2}} \right)= \\ $$$$\mathrm{3log}_{\mathrm{2}{x}} \left({x}^{\mathrm{3}} \right)\:\mathrm{is} \\ $$

Question Number 22215    Answers: 3   Comments: 0

solve the inequation −x^2 +3x−2>0

$${solve}\:{the}\:{inequation}\:−{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}>\mathrm{0} \\ $$

Question Number 22210    Answers: 1   Comments: 0

∫((4x)/e^(3x) )dx

$$\int\frac{\mathrm{4}{x}}{{e}^{\mathrm{3}{x}} }{dx} \\ $$

Question Number 22203    Answers: 1   Comments: 1

lim_(x→∞) e^x cos x

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{x}} \:\mathrm{cos}\:{x} \\ $$

Question Number 22221    Answers: 1   Comments: 1

∫ ((a_0 +b_0 x^2 )/((a+x)^2 ))dx

$$\int\:\:\frac{{a}_{\mathrm{0}} +{b}_{\mathrm{0}} {x}^{\mathrm{2}} }{\left({a}+{x}\right)^{\mathrm{2}} }{dx} \\ $$

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