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Question Number 209853 by peter frank last updated on 23/Jul/24

Commented by mr W last updated on 23/Jul/24

$t h e i r o n r o d m u s t b e o f a l e n g t h o f$ $44482 c m!$
	Answered by mr W last updated on 23/Jul/24

	$L_{B} = l e n g t h o f b r a s s b a r a t 10 ° C$ $L_{I} = l e n g t h o f i r o n b a r a t 10 ° C$ $L_{I} - L_{B} = 14 . . . (i)$ $a t 100 ° C :$ $l e n g t h o f b r a s s b a r :$ $L_{B 100} = L_{B} + L_{B} \times (100 - 10) \times 19 \times 10^{- 6}$ $l e n g t h o f i r o n b a r :$ $L_{I 100} = L_{I} + L_{I} \times (100 - 10) \times 12 \times 10^{- 6}$ $L_{B 100} - L_{I 100} = 14$ $L_{B} + L_{B} \times (100 - 10) \times 19 \times 10^{- 6} - L_{I} - L_{I} \times (100 - 10) \times 12 \times 10^{- 6} = 14 . . . (i i)$ $L_{B} \times 90 \times 19 - L_{I} \times 90 \times 12 = 28 \times 10^{6}$ $\Rightarrow (L_{I} - 14) \times 90 \times 19 - L_{I} \times 90 \times 12 = 28 \times 10^{6}$ $\Rightarrow L_{I} \approx 44482 c m, L_{B} \approx 44468 c m$ $c h e c k a t 100 ° C :$ $i r o n b a r : 44482 + 44482 \times 90 \times 12 \times 10^{- 6} = 44530 c m$ $b r a s s b a r : 44468 + 44468 \times 90 \times 19 \times 10^{- 6} = 44544 c m$
	Commented by mr W last updated on 23/Jul/24

	Commented by mr W last updated on 23/Jul/24

	$a t 10 ° C t h e d i f f e r e n c e i n t h e i r$ $l e n g t h e s i s 14 c m .$ $a t 100 ° C t h e d i f f e r e n c e i n t h e i r$ $l e n g t h e s s h o u l d a l s o b e 14 c m .$ $t h i s i s o n l y p o s s i b l e w h e n a t 10 ° C$ $t h e b r a s s r o d i s s h o r t e r t h a n t h e$ $i r o n r o d a n d a t 100 ° C t h e b r a s s r o d$ $i s l o n g e r t h a n t h e i r o n r o d . b u t t h e i r$ $d i f f e r e n c e i s i n b o t h c a s e s 14 c m .$
	Answered by Spillover last updated on 23/Jul/24

	$l_{B} = l e n g t h o f b r a s s l_{I} = l e n g t h o f i r o n$ $α_{B} = 19 \times 10^{- 6} / k α_{I} = 12 \times 10^{- 6} / k$ $l_{B} - l_{I} = 14 c m o r l_{I} - l_{B} = 14 c m$ $α_{B} = \frac{△ l_{B}}{l_{B} △ θ} l_{B} = \frac{△ l_{B}}{α_{B} △ θ}$ $α_{I} = \frac{△ l_{I}}{l_{I} △ θ} l_{I} = \frac{△ l_{I}}{α_{I} △ θ}$ $l_{I} - l_{B} = \frac{△ l_{I}}{α_{I} △ θ} - \frac{△ l_{B}}{α_{B} △ θ}$ $14 = \frac{△ l_{I}}{α_{I} △ θ} - \frac{△ l_{B}}{α_{B} △ θ} △ θ = 100 - 10 = 90 ° C$ $14 = \frac{△ l_{I}}{12 \times 10^{- 6} / k \times 90} - \frac{△ l_{B}}{19 \times 10^{- 6} / k \times 90}$ $14 = 925.93 △ l_{I} - 584.8 △ l_{B}$ $N . B$ $f o r t h e d i f f e r e n c e t o r e m a i n t h e s a m e$ $△ l_{I} m u s t b e e q u a l t o △ l_{B} [△ l_{I} = △ l_{B}]$ $14 = 925.93 △ l_{I} - 584.8 △ l_{B}$ $14 = 925.93 △ l_{I} - 584.8 △ l_{I}$ $14 = 341.13 △ l_{I}$ $△ l_{I} = 0.041 c m$ $α_{I} = \frac{△ l_{I}}{l_{I} △ θ}$ $l_{I} = \frac{△ l_{I}}{α_{I} △ θ} = \frac{0.041}{12 \times 10^{- 6} \times 90} = 37.96$ $l_{I} = 37.96 \approx 38$ $L e n g t h o f t h e i r o n m u s t b e 38 c m$
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