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A glass bottle full of mercury has mass 500g. On being heated through 35°C, 2.43g of mercury are expelled. calculate the mass of mercury remaining in the bottle (Cubic expansivity of mercury is 1.8 × 10^(−4) per K. linear expansivity of glass is 8.0 × 10^(−6) per K.

$A glass bottle full of mercury has mass 500 g . On being heated through 35 ° C,$ $2 . 43 g of mercury are expelled . calculate the mass of mercury remaining in$ $the bottle (Cubic expansivity of mercury is 1.8 \times 10^{- 4} per K .$ $linear expansivity of glass is 8.0 \times 10^{- 6} per K .$

Question Number 27688 Answers: 0 Comments: 0

Write the series,indicating the 5th term,the 5th partial sum 0+1+3+...+(((n^2 +n)/2))+...

$W r i t e t h e s e r i e s, i n d i c a t i n g t h e$ $5 t h t e r m, t h e 5 t h p a r t i a l s u m$ $0 + 1 + 3 + . . . + (\frac{n^{2} + n}{2}) + . . .$

Question Number 27685 Answers: 0 Comments: 2

Write the first five series indicating the 5th term,5th partial sum Σ_(n=1) ^∞ t_n , where t_n = { ((1 for n=1)),(((1/2) for n=2)),((1−(1/2)+...+(−1)^(n+1) ((1/n)) for n>2)) :}

$W r i t e t h e f i r s t f i v e s e r i e s i n d i c a t i n g$ $t h e 5 t h t e r m, 5 t h p a r t i a l s u m$ $\underset{n = 1}{\sum^{\infty}} t_{n}, w h e r e$ $t_{n} = {\begin{cases} 1 f o r n = 1 \\ \frac{1}{2} f o r n = 2 \\ 1 - \frac{1}{2} + . . . + {(- 1)}^{n + 1} (\frac{1}{n}) f o r n > 2 \end{cases}$

Question Number 27609 Answers: 0 Comments: 0

Δ=(√(m×φ)) Δ=mass gap m=mass φ=phi calculate phi to the same number of decimal places as the mass. use the mass of an electron

$Δ = \sqrt{m \times ϕ}$ $Δ = m a s s g a p$ $m = m a s s$ $ϕ = p h i$ $c a l c u l a t e p h i t o t h e s a m e n u m b e r o f$ $d e c i m a l p l a c e s a s t h e m a s s .$ $u s e t h e m a s s o f a n e l e c t r o n$

Question Number 27599 Answers: 0 Comments: 0

let give the equation x^6 −x−1=0 by using Newton methodfind the approximate value of the real?root for this equation.

$l e t g i v e t h e e q u a t i o n x^{6} - x - 1 = 0 b y u s i n g N e w t o n m e t h o d f i n d$ $t h e a p p r o x i m a t e v a l u e o f t h e r e a l ? r o o t f o r t h i s e q u a t i o n .$

Question Number 27603 Answers: 0 Comments: 1

Find the value of i^i ?

$F i n d t h e v a l u e o f i^{i} ?$

Question Number 27536 Answers: 0 Comments: 0

m_1 s_1 (x−𝛉)=m_2 s_2 (𝛉−y) ; x=? ;y=? 𝛉=? solve it as an equation....

$m_{1} s_{1} (x - θ) = m_{2} s_{2} (θ - y); x = ?; y = ? θ = ?$ $solve it as an equation . . . .$

Question Number 27520 Answers: 0 Comments: 0

Question Number 27430 Answers: 0 Comments: 0

A 2000kg space capsule is traveling away from the earth, determine the gravitational field strenght and gravitational force on the capsule due to the earth when it is (a) At a distance from the earth′s surface equal to the radius of the earth (b) At a very large distance away from the earth (Take g = 9.8Nkg^(−1) on earth surface)

$A 2000 kg space capsule is traveling away from the earth, determine the$ $gravitational field strenght and gravitational force on the capsule due to the$ $earth when it is$ $(a) At a distance from the {earth}^{'} s surface equal to the radius of the earth$ $(b) At a very large distance away from the earth (Take g = 9 . {8 Nkg}^{- 1} on$ $earth surface)$

Question Number 27427 Answers: 0 Comments: 0

A particle of mass 2kg moves in a force field depending on a time t given by F = 24t^2 i + (36t − 16)j − 12tk assuming that at t = 0 the particle is located at r_0 = 3i − j + 4k and has v_0 = 6i + 5j − 8k. Find (a) Velocity at any time t (b) Position at any time t (c) τ (torgue) at any time t (d) Angular momentum at any time t above the Origin

$A particle of mass 2 kg moves in a force field depending on a time t given by$ $F = {24 t}^{2} i + (36 t - 16) j - 12 tk assuming that at t = 0 the particle is located$ $at r_{0} = 3 i - j + 4 k and has v_{0} = 6 i + 5 j - 8 k . Find$ $(a) Velocity at any time t$ $(b) Position at any time t$ $(c) τ (torgue) at any time t$ $(d) Angular momentum at any time t above the Origin$

Question Number 27428 Answers: 0 Comments: 0

Find the workdone in moving an object along a vector r = 3i + 2j − 5k if the applied force is F = 2i − j − k

$Find the workdone in moving an object along a vector$ $r = 3 i + 2 j - 5 k if the applied force is F = 2 i - j - k$

Question Number 27399 Answers: 1 Comments: 0

A and B are walking along a circular track.They start from same point at 8:00 am. A can walk 2 rounds per hour and B can walk 3 rounds per hour. How many times they cross each other before 9:30 am if they walk (i) Opposite to each other. (ii) In same direction. ?

$A and B are walking along a$ $circular track . They start from$ $same point at 8 : 00 am .$ $A can walk 2 rounds per hour$ $and B can walk 3 rounds per hour .$ $How many times they cross each$ $other before 9 : 30 am if they walk$ $(i) Opposite to each other .$ $(ii) In same direction .$ $?$

Question Number 27335 Answers: 0 Comments: 1

if 2 chords of ellipse have the same distance from the centre of ellipse and the eccentric angle of the end points of the chords are respectivly α β γ δ then prove that tan (α/2)×tan (β/2)×tan (γ/2)×tan (δ/2)=1

$i f 2 c h o r d s o f e l l i p s e h a v e t h e s a m e$ $d i s t a n c e f r o m t h e c e n t r e o f e l l i p s e$ $a n d t h e e c c e n t r i c a n g l e o f t h e e n d p o i n t s o f t h e c h o r d s$ $a r e r e s p e c t i v l y α β γ δ t h e n p r o v e t h a t$ $\tan \frac{α}{2} \times \tan \frac{β}{2} \times \tan \frac{γ}{2} \times \tan \frac{δ}{2} = 1$