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AllQuestion and Answers: Page 1822

Question Number 23721 Answers: 1 Comments: 0

If symbols have their usual meaning then (1/r^2 ) + (1/r_1 ^2 ) + (1/r_2 ^2 ) + (1/r_3 ^2 ) = (1) ((a^2 + b^2 + c^2 )/s^2 ) (2) (Δ/(a^2 + b^2 + c^2 )) (3) ((a^2 + b^2 + c^2 )/Δ^2 ) (4) ((a + b + c)/Δ^2 )

$$\mathrm{If}\:\mathrm{symbols}\:\mathrm{have}\:\mathrm{their}\:\mathrm{usual}\:\mathrm{meaning} \\ $$$$\mathrm{then}\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}_{\mathrm{2}} ^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}_{\mathrm{3}} ^{\mathrm{2}} }\:= \\ $$$$\left(\mathrm{1}\right)\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{s}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{\Delta}{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{\Delta^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{{a}\:+\:{b}\:+\:{c}}{\Delta^{\mathrm{2}} } \\ $$

Question Number 23722 Answers: 0 Comments: 4

A block of mass m is pulled on the smooth horizontal floor using two methods I and II. The ratio of acceleration (a_I /a_(II) ) is

$$\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{pulled}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{floor}\:\mathrm{using}\:\mathrm{two} \\ $$$$\mathrm{methods}\:\mathrm{I}\:\mathrm{and}\:\mathrm{II}.\:\mathrm{The}\:\mathrm{ratio}\:\mathrm{of} \\ $$$$\mathrm{acceleration}\:\frac{{a}_{{I}} }{{a}_{{II}} }\:\mathrm{is} \\ $$

Question Number 18004 Answers: 1 Comments: 1

Question Number 18003 Answers: 1 Comments: 0

The value of cosA∙cos2A∙cos2^2 A ..... cos(2^(n − 1) A), where A ∈ R may be (1) 1 (2) 2 (3) −1 (4) ((sin 2^n A)/(2^n sin A))

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}{A}\centerdot\mathrm{cos2}{A}\centerdot\mathrm{cos2}^{\mathrm{2}} {A}\:.....\:\mathrm{cos}\left(\mathrm{2}^{{n}\:−\:\mathrm{1}} {A}\right), \\ $$$$\mathrm{where}\:{A}\:\in\:{R}\:\mathrm{may}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:−\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{sin}\:\mathrm{2}^{{n}} \:{A}}{\mathrm{2}^{{n}} \:\mathrm{sin}\:{A}} \\ $$

Question Number 17999 Answers: 1 Comments: 0

A large number of bullets are fired in all direction with same speed u. The maximum area on the ground covered by these bullets will be (1) π.(u^2 /g) (2) π.(u^4 /g^2 ) (3) (π/4).(u^4 /g^2 ) (4) (π/2).(u^4 /g^2 )

$$\mathrm{A}\:\mathrm{large}\:\mathrm{number}\:\mathrm{of}\:\mathrm{bullets}\:\mathrm{are}\:\mathrm{fired}\:\mathrm{in} \\ $$$$\mathrm{all}\:\mathrm{direction}\:\mathrm{with}\:\mathrm{same}\:\mathrm{speed}\:{u}.\:\mathrm{The} \\ $$$$\mathrm{maximum}\:\mathrm{area}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{covered} \\ $$$$\mathrm{by}\:\mathrm{these}\:\mathrm{bullets}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\pi.\frac{{u}^{\mathrm{2}} }{{g}} \\ $$$$\left(\mathrm{2}\right)\:\pi.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\pi}{\mathrm{4}}.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\pi}{\mathrm{2}}.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$

Question Number 17997 Answers: 0 Comments: 0

Solve on Z_(18) { (([6]_(18) x+[7]_(18) y=[1]_(18) )),(([2]_(18) x+[3]_(18) y=[11]_(18) )) :}

$${Solve}\:{on}\:\mathbb{Z}_{\mathrm{18}} \\ $$$$\begin{cases}{\left[\mathrm{6}\right]_{\mathrm{18}} {x}+\left[\mathrm{7}\right]_{\mathrm{18}} {y}=\left[\mathrm{1}\right]_{\mathrm{18}} }\\{\left[\mathrm{2}\right]_{\mathrm{18}} {x}+\left[\mathrm{3}\right]_{\mathrm{18}} {y}=\left[\mathrm{11}\right]_{\mathrm{18}} }\end{cases} \\ $$

Question Number 17995 Answers: 0 Comments: 0

A pendulum bob of mass 2kg is attached to a string 2m long and made to revolve in horizontal circle of radius 0.8 , find the tension in the string.

$$\mathrm{A}\:\mathrm{pendulum}\:\mathrm{bob}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a}\:\mathrm{string}\:\:\mathrm{2m}\:\mathrm{long}\:\mathrm{and}\:\mathrm{made}\:\mathrm{to} \\ $$$$\mathrm{revolve}\:\mathrm{in}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{0}.\mathrm{8}\:,\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}. \\ $$

Question Number 17991 Answers: 2 Comments: 3

Evaluate (√(1+2(√(1+3(√(1+4(√(1+...))))))))

$${Evaluate}\:\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+...}}}} \\ $$

Question Number 17989 Answers: 0 Comments: 8

Solve for x x^x^x^x^(....) = 4

$${Solve}\:{for}\:{x} \\ $$$${x}^{{x}^{{x}^{{x}^{....} } } } =\:\mathrm{4} \\ $$

Question Number 17985 Answers: 1 Comments: 0

solve simultaneously. x^3 + y^3 = 35 x^4 + y^4 = 97

$$\mathrm{solve}\:\mathrm{simultaneously}. \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{35} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{4}} \:=\:\mathrm{97} \\ $$

Question Number 17983 Answers: 1 Comments: 0

Evaluate cos(20°)cos(40°)cos(80°). This question is just for fun and practice. Evryone who wants can answer this question.

$${Evaluate}\:{cos}\left(\mathrm{20}°\right){cos}\left(\mathrm{40}°\right){cos}\left(\mathrm{80}°\right). \\ $$$${This}\:{question}\:{is}\:{just}\:{for}\:{fun}\:{and}\:{practice}. \\ $$$${Evryone}\:{who}\:{wants}\:{can}\:{answer}\:{this}\:{question}. \\ $$

Question Number 17976 Answers: 1 Comments: 0

A driver is 5.2 below the surface of water of density 1000kg/m. if the atmospheric pressure is 1.02 × 10^5 Pa. calculate the pressure on the driver.

$$\mathrm{A}\:\mathrm{driver}\:\mathrm{is}\:\mathrm{5}.\mathrm{2}\:\mathrm{below}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{water}\:\mathrm{of}\:\mathrm{density}\:\mathrm{1000kg}/\mathrm{m}.\:\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{atmospheric}\:\mathrm{pressure}\:\mathrm{is}\:\mathrm{1}.\mathrm{02}\:×\:\mathrm{10}^{\mathrm{5}} \:\mathrm{Pa}.\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{pressure}\:\mathrm{on}\:\mathrm{the}\:\mathrm{driver}. \\ $$

Question Number 17970 Answers: 1 Comments: 0

Question Number 17969 Answers: 0 Comments: 0

Question Number 17963 Answers: 0 Comments: 4

Question Number 17961 Answers: 1 Comments: 0

Question Number 17959 Answers: 1 Comments: 0

Question Number 17957 Answers: 1 Comments: 0

prove that (1+(a/x))^n =(1+((an)/x)) , x≫a

$$\mathrm{prove}\:\mathrm{that}\:\left(\mathrm{1}+\frac{\mathrm{a}}{\mathrm{x}}\right)^{\mathrm{n}} =\left(\mathrm{1}+\frac{\mathrm{an}}{\mathrm{x}}\right)\:,\:\:\mathrm{x}\gg\mathrm{a} \\ $$

Question Number 17939 Answers: 1 Comments: 1

∫secxdx

$$\int{secxdx}\: \\ $$

Question Number 17938 Answers: 2 Comments: 2

If only downward motion along lines is allowed, what is the total number of paths from point P to point Q in the figure below?

$$\mathrm{If}\:\mathrm{only}\:\mathrm{downward}\:\mathrm{motion}\:\mathrm{along}\:\mathrm{lines}\:\mathrm{is} \\ $$$$\mathrm{allowed},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{paths}\:\mathrm{from}\:\mathrm{point}\:\mathrm{P}\:\mathrm{to}\:\mathrm{point}\:\mathrm{Q}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{figure}\:\mathrm{below}? \\ $$

Question Number 17936 Answers: 2 Comments: 1

Two particles start moving towards each other with constant acceleration of 1 m/s^2 . If their initial separation is 100 m, find after what time they will meet each other (A) 40s (B) 45s (C) 50 s (D)55s

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{start}\:\mathrm{moving}\:\mathrm{towards} \\ $$$$\mathrm{each}\:\mathrm{other}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\:\mathrm{1}\:{m}/{s}^{\mathrm{2}} .\:\mathrm{If}\:\mathrm{their}\:\mathrm{initial}\:\mathrm{separation}\:\mathrm{is} \\ $$$$\mathrm{100}\:\mathrm{m},\:\mathrm{find}\:\mathrm{after}\:\mathrm{what}\:\mathrm{time}\:\mathrm{they}\:\mathrm{will} \\ $$$$\mathrm{meet}\:\mathrm{each}\:\mathrm{other} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{40}{s}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{45}{s}\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{50}\:{s}\:\left(\mathrm{D}\right)\mathrm{55}{s} \\ $$

Question Number 17919 Answers: 0 Comments: 0

How do the electronic configurations of the elements with Z = 107 − 109 differ from one another?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{the}\:\mathrm{electronic}\:\mathrm{configurations}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{elements}\:\mathrm{with}\:\mathrm{Z}\:=\:\mathrm{107}\:−\:\mathrm{109}\:\mathrm{differ} \\ $$$$\mathrm{from}\:\mathrm{one}\:\mathrm{another}? \\ $$

Question Number 17918 Answers: 0 Comments: 0

Write the electronic configuration and the block to which an element with Z = 90 belongs.

$$\mathrm{Write}\:\mathrm{the}\:\mathrm{electronic}\:\mathrm{configuration}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{to}\:\mathrm{which}\:\mathrm{an}\:\mathrm{element}\:\mathrm{with} \\ $$$$\mathrm{Z}\:=\:\mathrm{90}\:\mathrm{belongs}. \\ $$

Question Number 17901 Answers: 1 Comments: 2

Question Number 17895 Answers: 0 Comments: 2

Question Number 17909 Answers: 0 Comments: 1

Please answer Q. 17525

$${Please}\:{answer}\:{Q}.\:\mathrm{17525} \\ $$

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