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Question Number 22929 Answers: 1 Comments: 0

Question Number 22928 Answers: 1 Comments: 0

Question Number 22956 Answers: 1 Comments: 4

Question Number 22923 Answers: 1 Comments: 0

Question Number 22922 Answers: 0 Comments: 0

Question Number 22896 Answers: 0 Comments: 1

how can demonstred 17^(4n+1) +3×9^(2n+1) ≡0[11]

$h o w c a n d e m o n s t r e d$ $17^{4 n + 1} + 3 \times 9^{2 n + 1} \equiv 0 [11]$

Question Number 22941 Answers: 1 Comments: 3

Question Number 22893 Answers: 1 Comments: 0

A wedge has two equally rough faces each inclined at 30° to the horizontal.Masses of 5kg and 2kg ,one on each face,are connected by a light string passing over a smooth pulley at the top of the wedge.The coefficient of friction μ, between each masses and the surface of the wedge is 0.2.Find the acceleration of the masses when they are released.

$A w e d g e h a s t w o e q u a l l y r o u g h$ $f a c e s e a c h i n c l i n e d a t 30 ° t o t h e$ $h o r i z o n t a l . M a s s e s o f 5 k g a n d 2 k g$ $, o n e o n e a c h f a c e, a r e c o n n e c t e d$ $b y a l i g h t s t r i n g p a s s i n g o v e r a$ $s m o o t h p u l l e y a t t h e t o p o f t h e$ $w e d g e . T h e c o e f f i c i e n t o f f r i c t i o n$ $μ, b e t w e e n e a c h m a s s e s a n d t h e$ $s u r f a c e o f t h e w e d g e i s 0.2 . F i n d$ $t h e a c c e l e r a t i o n o f t h e m a s s e s$ $w h e n t h e y a r e r e l e a s e d .$

Question Number 22892 Answers: 0 Comments: 3

A horizontal force of 10N just prevents a mass of 2kg from sliding down a rough plane inclined at 45° to the horizontal. Find the coefficient of friction.

$A h o r i z o n t a l f o r c e o f 10 N j u s t$ $p r e v e n t s a m a s s o f 2 k g f r o m$ $s l i d i n g d o w n a r o u g h p l a n e$ $i n c l i n e d a t 45 ° t o t h e h o r i z o n t a l .$ $F i n d t h e c o e f f i c i e n t o f f r i c t i o n .$

Question Number 22883 Answers: 1 Comments: 1

The correct order of second ionisation energy in the following : (a) F > O > N > C (b) O > F > N > C (c) O > N > F > C (d) C > N > O > F

$The correct order of second ionisation$ $energy in the following :$ $(a) F > O > N > C$ $(b) O > F > N > C$ $(c) O > N > F > C$ $(d) C > N > O > F$

Question Number 22881 Answers: 0 Comments: 3

which is your favourite subject?

$w h i c h i s y o u r f a v o u r i t e s u b j e c t ?$

Question Number 22879 Answers: 1 Comments: 0

The Enthalpy of neutralization of acetic acid and sodium hydroxide is −55.4 kJ. What is the enthalpy of ionisation of acetic acid?

$The Enthalpy of neutralization of$ $acetic acid and sodium hydroxide is$ $- 55.4 kJ . What is the enthalpy of$ $ionisation of acetic acid ?$

Question Number 22872 Answers: 1 Comments: 4

Question Number 22856 Answers: 0 Comments: 1

solve: sin(x)=2, x∈C

$solve : \sin (x) = 2, x \in C$

Question Number 22855 Answers: 1 Comments: 1

Equal forces act on two bodies which are initially at rest,for equal times.If the ratio of the masses is 4:3,What is the ratio of the distance covered?

$E q u a l f o r c e s a c t o n t w o b o d i e s$ $w h i c h a r e i n i t i a l l y a t r e s t, f o r$ $e q u a l t i m e s . I f t h e r a t i o o f t h e$ $m a s s e s i s 4 : 3, W h a t i s t h e r a t i o$ $o f t h e d i s t a n c e c o v e r e d ?$

Question Number 22884 Answers: 1 Comments: 0

A body of mass 0.3 kg is taken up an inclined plane of length 10 m and height 5 m, and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the kinetic energy of the body at the end of the trip?

$A body of mass 0.3 kg is taken up an$ $inclined plane of length 10 m and$ $height 5 m, and then allowed to slide$ $down to the bottom again . The$ $coefficient of friction between the body$ $and the plane is 0.15 . What is the$ $kinetic energy of the body at the end of$ $the trip ?$

Question Number 22837 Answers: 0 Comments: 0

The effective nuclear charge of N on its last electron is:−

$The effective nuclear charge of N on its$ $last electron is : -$

Question Number 22871 Answers: 1 Comments: 0

If f(n)=(1/n)[(n+1)(n+2)(n+3)...(n+n)]^(1/n) , then lim_(n→∞) f(n) =

$If f (n) = \frac{1}{n} {[(n + 1) (n + 2) (n + 3) . . . (n + n)]}^{1 / n},$ $then \lim_{n \to \infty} f (n) =$

Question Number 22843 Answers: 1 Comments: 0

calculate the value of ∫_0 ^(1000) e^(x−[x]) dx =? where.... [x] is greatest integer functon

$c a l c u l a t e t h e v a l u e o f \int_{0}^{1000} e^{x - [x]} d x = ?$ $w h e r e . . . . [x] i s g r e a t e s t i n t e g e r f u n c t o n$

Question Number 22820 Answers: 0 Comments: 2

total number of permutations of five abjects → A,A,A,B,B in a circle?

$t o t a l n u m b e r o f p e r m u t a t i o n s o f$ $f i v e a b j e c t s \to A, A, A, B, B$ $i n a c i r c l e ?$

Question Number 22874 Answers: 4 Comments: 0

Question Number 22813 Answers: 1 Comments: 0

The nucleus Fe^(57) emits a γ-ray of energy 14.4 keV. If the mass of the nucleus is 56.935 u, calculate the recoil energy of nucleus.

$The nucleus {Fe}^{57} emits a γ - ray of energy$ $14.4 keV . If the mass of the nucleus is$ $56.935 u, calculate the recoil energy of$ $nucleus .$

Question Number 22797 Answers: 1 Comments: 0

He^+ in ground state is hit by photon thus exciting electron to second excited state. The energy absorbed in this process is (approx) (1) 2.18 × 10^(−14) J (2) 7.75 × 10^(−18) J (3) 4.5 × 10^(−18) J (4) 5 × 10^(−18) J

${He}^{+} in ground state is hit by photon$ $thus exciting electron to second excited$ $state . The energy absorbed in this$ $process is (approx)$ $(1) 2.18 \times 10^{- 14} J$ $(2) 7.75 \times 10^{- 18} J$ $(3) 4.5 \times 10^{- 18} J$ $(4) 5 \times 10^{- 18} J$

Question Number 22787 Answers: 0 Comments: 2

sin^(−1) (sin 10)=10 or 3π−10 Ans is 3π−10 How

$\sin^{- 1} (\sin 10) = 10 or 3 π - 10$ $Ans is 3 π - 10 How$

Question Number 22782 Answers: 0 Comments: 6

Successive ionisation energies of element is 738, 1450, 10540, 13630 kJ/mole respectively. The element is (1) Na (2) Mg (3) Al (4) K

$Successive ionisation energies of$ $element is 738, 1450, 10540, 13630$ $kJ / mole respectively . The element is$ $(1) Na$ $(2) Mg$ $(3) Al$ $(4) K$

Question Number 22778 Answers: 0 Comments: 2

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