B\u00e0i 36 : S\u1ef1 n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a v\u1eadt r\u1eafn <\/h2>\n\n\n\n

<\/figure>\n\n\n\n

C1.<\/strong> ( trang 188 sgk V\u1eadt L\u00fd 10): T\u00ednh h\u1ec7 s\u1ed1 \n c\u1ee7a m\u1ed7i l\u1ea7n \u0111o ghi trong b\u1ea3ng 36.1. X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a h\u1ec7 s\u1ed1 \u03b1.<\/p>\n\n\n\n

V\u1edbi sai s\u1ed1 kho\u1ea3ng 5%, nh\u1eadn x\u00e9t xem h\u1ec7 s\u1ed1 \u03b1 c\u00f3 gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i hay thay \u0111\u1ed5i?<\/ins><\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\n

\n\t\t\tNhi\u1ec7t \u0111\u1ed9 ban \u0111\u1ea7u: t0<\/sub> = 20\u00baC\n\t\t\t\u0110\u1ed9 d\u00e0i ban \u0111\u1ea7u: l0<\/sub> = 500 mm\n\t\t<\/td><\/tr>
\u2206t (\u00baC)<\/td>	\u2206l (mm)<\/td>	\n\t\t<\/td><\/tr>
30<\/td>	0,25<\/td>	1,67.10-5<\/sup><\/td><\/tr>
40<\/td>	0,33<\/td>	1,65.10-5<\/sup><\/td><\/tr>
50<\/td>	0,41<\/td>	1,64.10-5<\/sup><\/td><\/tr>
60<\/td>	0,49<\/td>	1,63.10-5<\/sup><\/td><\/tr>
70<\/td>	0,58<\/td>	1,66.10-5<\/sup><\/td><\/tr><\/tbody><\/table>\n\n\n\n + Gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a h\u1ec7 s\u1ed1 \u03b1: <\/p>\n\n\n\n <\/figure>\n\n\n\n V\u1edbi sai s\u1ed1 5%, h\u1ec7 s\u1ed1 \u03b1 coi nh\u01b0 c\u00f3 gi\u00e1 tr\u1ecb kh\u00f4ng thay \u0111\u1ed5i v\u00e0 \u0111\u01b0\u1ee3c vi\u1ebft d\u01b0\u1edbi d\u1ea1ng:<\/p>\n\n\n\n <\/figure>\n\n\n\n <\/ins><\/p>\n\n\n\n <\/figure>\n\n\n\n C2.<\/strong> ( trang 189 sgk V\u1eadt L\u00fd 10): D\u1ef1a v\u00e0o c\u00f4ng th\u1ee9c \n, h\u00e3y cho bi\u1ebft \u00fd ngh\u0129a c\u1ee7a h\u1ec7 s\u1ed1 n\u1edf d\u00e0i \u03b1.<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\n <\/figure>\n\n\n\n <\/figure>\n\n\n\n g\u1ecdi l\u00e0 \u0111\u1ed9 d\u00e3n t\u1ec9 \u0111\u1ed1i c\u1ee7a thanh r\u1eafn khi nhi\u1ec7t \u0111\u1ed9 t\u0103ng. T\u1eeb c\u00f4ng th\u1ee9c ta th\u1ea5y khi \u0394t = 1\u00baC th\u00ec , t\u1ee9c h\u1ec7 s\u1ed1 n\u1edf d\u00e0i c\u00f3 tr\u1ecb s\u1ed1 b\u1eb1ng \u0111\u1ed9 d\u00e3n t\u1ec9 \u0111\u1ed1i khi nhi\u1ec7t \u0111\u1ed9 t\u0103ng 1 \u0111\u1ed9.<\/p>\n\n\n\n B\u00e0i 4 (trang 197 SGK V\u1eadt L\u00fd 10) :<\/strong> T\u1ea1i sao khi \u0111\u1ed5 n\u01b0\u1edbc s\u00f4i v\u00e0o trong c\u1ed1c th\u1ee7y tinh th\u00ec c\u1ed1c th\u1ee7y tinh hay b\u1ecb n\u1ee9t v\u1ee1, c\u00f2n c\u1ed1c th\u1ea1ch anh kh\u00f4ng b\u1ecb n\u1ee9t v\u1ee1?<\/ins><\/p>\n\n\n\n A. V\u00ec c\u1ed1c th\u1ea1ch anh c\u00f3 th\u00e0nh d\u00e0y h\u01a1n<\/p>\n\n\n\n B. V\u00ec c\u1ed1c th\u1ea1ch anh c\u00f3 \u0111\u00e1y d\u00e0y h\u01a1n<\/p>\n\n\n\n C. V\u00ec th\u1ea1ch anh c\u1ee9ng h\u01a1n th\u1ee7y tinh<\/p>\n\n\n\n D. V\u00ec th\u1ea1ch anh c\u00f3 h\u1ec7 s\u1ed1 n\u1edf kh\u1ed1i nh\u1ecf h\u01a1n th\u1ee7y tinh.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n Ch\u1ecdn D.<\/p>\n\n\n\n Th\u1ea1ch anh c\u00f3 h\u1ec7 s\u1ed1 n\u1edf d\u00e0i \u03b11<\/sub> = 0,6.10-6<\/sup> K-1<\/sup> nh\u1ecf h\u01a1n h\u1ec7 s\u1ed1 n\u1edf d\u00e0i c\u1ee7a th\u1ee7y tinh \u03b12<\/sub> = 9.10-6<\/sup> K-1<\/sup>, n\u00ean khi g\u1eb7p nhi\u1ec7t th\u00ec l\u1edbp th\u1ee7y tinh m\u1eb7t trong c\u1ed1c gi\u00e3n n\u1edf nhanh h\u01a1n so v\u1edbi b\u00ean ngo\u00e0i, g\u00e2y bi\u1ebfn d\u1ea1ng \u0111\u1ed9t ng\u1ed9t n\u00ean d\u1ec5 v\u1ee1. C\u00f2n th\u1ea1ch anh gi\u00e3n n\u1ee1 ch\u1eadm n\u00ean b\u00ean trong c\u1ed1c v\u00e0 b\u00ean ngo\u00e0i gi\u00e3n n\u1ee1 g\u1ea7n nh\u01b0 nhau n\u00ean kh\u00f4ng g\u00e2y bi\u1ebfn d\u1ea1ng \u0111\u1ed9t ng\u1ed9t, c\u1ed1c kh\u00f4ng b\u1ecb n\u1ee9t v\u1ee1.<\/p>\n\n\n\n B\u00e0i 5 (trang 197 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed9t th\u01b0\u1edbc th\u00e9p \u1edf 20o<\/sup> C c\u00f3 \u0111\u1ed9 d\u00e0i 1000 mm. Khi nhi\u1ec7t \u0111\u1ed9 t\u0103ng \u0111\u1ebfn 40o<\/sup> C, th\u01b0\u1edbc th\u00e9p n\u00e0y d\u00e0i th\u00eam bao nhi\u00eau?<\/ins><\/p>\n\n\n\n A. 2,4 mm ;         B. 3,2 mm<\/p>\n\n\n\n C. 0,22 mm ;         D. 4,2 mm<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n – Ch\u1ecdn C<\/strong>.<\/p>\n\n\n\n – \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c \u0394l = l – lo<\/sub> = \u03b1lo<\/sub>\u0394t, ta \u0111\u01b0\u1ee3c:<\/p>\n\n\n\n \u0394l = 11.10-6<\/sup>.1.(40 – 20) = 220.10-6<\/sup> (m) = 0,22 mm<\/p>\n\n\n\n B\u00e0i 6 (trang 197 SGK V\u1eadt L\u00fd 10) :<\/strong> Kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a s\u1eaft \u1edf 800o<\/sup> C b\u1eb1ng bao nhi\u00eau? Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a n\u00f3 \u1edf 0o<\/sup> C l\u00e0 7,800.103<\/sup> kg\/m3<\/sup><\/ins><\/p>\n\n\n\n A. 7,900.103<\/sup> kg\/m3<\/sup> ;         B. 7,599.103<\/sup> kg\/m3<\/sup><\/p>\n\n\n\n C. 7,857.103<\/sup> kg\/m3<\/sup> ;         D. 7,485.103<\/sup> kg\/m3<\/sup><\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n – Ch\u1ecdn B<\/strong>.<\/p>\n\n\n\n Ta c\u00f3: kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a m\u1ed9t ch\u1ea5t \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng:<\/p>\n\n\n\n <\/figure>\n\n\n\n M\u1eb7t kh\u00e1c, ta c\u00f3:<\/p>\n\n\n\n <\/figure>\n\n\n\n Thay s\u1ed1 v\u1edbi h\u1ec7 s\u1ed1 n\u1edf d\u00e0i c\u1ee7a s\u1eaft \u03b1 = 11.10-6<\/sup> K-1<\/sup> ta \u0111\u01b0\u1ee3c:<\/p>\n\n\n\n <\/figure>\n\n\n\n B\u00e0i 7 (trang 197 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed9t d\u00e2y t\u1ea3i \u0111i\u1ec7n \u1edf 20o<\/sup>C c\u00f3 \u0111\u1ed9 d\u00e0i 1 800 m. H\u00e3y x\u00e1c \u0111\u1ecbnh \u0111\u1ed9 n\u1edf d\u00e0i c\u1ee7a d\u00e2y t\u1ea3i \u0111i\u1ec7n n\u00e0y khi nhi\u1ec7t \u0111\u1ed9 t\u0103ng l\u00ean \u0111\u1ebfn 50o<\/sup> C v\u1ec1 m\u00f9a h\u00e8. Cho bi\u1ebft h\u1ec7 s\u1ed1 n\u1edf d\u00e0i c\u1ee7a d\u00e2y t\u1ea3i \u0111i\u1ec7n l\u00e0 \u03b1 = 11,5.10(-6)<\/sup> K(-1)<\/sup><\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n t1<\/sub> = 20o<\/sup> C, l1<\/sub> = 1800 m<\/p>\n\n\n\n t2<\/sub> = 50o<\/sup> C<\/p>\n\n\n\n \u03b1 = 11,5.10-6<\/sup> (k-1<\/sup>)<\/p>\n\n\n\n \u0394l = ?<\/p>\n\n\n\n \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c :<\/p>\n\n\n\n \u0394l = \u03b1l1<\/sub>\u0394t<\/p>\n\n\n\n \u0394l = 11,5.10-6<\/sup>.1800.(50 – 20) = 0,621 m<\/p>\n\n\n\n V\u1eady \u0111\u1ed9 n\u1edf d\u00e0i c\u1ee7a d\u00e2y t\u1ea3i \u0111i\u1ec7n l\u00e0 \u0394l = 0,621 (m)<\/p>\n\n\n\n B\u00e0i 8 (trang 197 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed7i thanh ray c\u1ee7a \u0111\u01b0\u1eddng s\u1eaft \u1edf nhi\u1ec7t \u0111\u1ed9 15o<\/sup>\n C c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 12,5 m. N\u1ebfu hai \u0111\u1ea7u c\u00e1c thanh ray khi \u0111\u00f3 ch\u1ec9 \u0111\u1eb7t c\u00e1ch \nnhau 4,50 mm, th\u00ec c\u00e1c thanh ray n\u00e0y c\u00f3 th\u1ec3 ch\u1ecbu \u0111\u01b0\u1ee3c nhi\u1ec7t \u0111\u1ed9 l\u1edbn nh\u1ea5t \nb\u1eb1ng bao nhi\u00eau \u0111\u1ec3 ch\u00fang kh\u00f4ng b\u1ecb u\u1ed1n cong do t\u00e1c d\u1ee5ng n\u1edf v\u00ec nhi\u1ec7t? Cho \nbi\u1ebft h\u1ec7 s\u1ed1 n\u1edf d\u00e0i c\u1ee7a m\u1ed7i thanh ray l\u00e0 \u03b1 = 12. 10-6<\/sup> (k-1<\/sup>).<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n     t1<\/sub> = 15o<\/sup>C<\/p>\n\n\n\n     l1<\/sub> = 12,5 m<\/p>\n\n\n\n     \u0394l = 4,5 mm = 4,5.10-3 m<\/p>\n\n\n\n     \u03b1 = 12.10-6<\/sup> K-1<\/sup><\/p>\n\n\n\n     t = ?<\/p>\n\n\n\n Kho\u1ea3ng c\u00e1ch gi\u1eefa hai thanh ray li\u00ean ti\u1ebfp nhau ch\u00ednh l\u00e0 \u0111\u1ed9 n\u1edf d\u00e0i c\u1ee7a m\u1ed7i thanh khi thnah \u0111\u1ea1t \u0111\u1ebfn nhi\u1ec7t \u0111\u1ed9 l\u1edbn nh\u1ea5t t\u00baC.<\/p>\n\n\n\n Ta c\u00f3: \u0394l = \u03b1.l0.\u0394t<\/p>\n\n\n\n <\/figure>\n\n\n\n \u2192 \u0110\u1ed9 t\u0103ng nhi\u1ec7t \u0111\u1ed9 t\u1ed1i \u0111a l\u00e0: <\/p>\n\n\n\n M\u00e0 \u0394t = t \u2013 t0<\/sub> \u21d2 t = \u0394t + t0<\/sub> = 45\u00ba<\/p>\n\n\n\n V\u1eady thanh ray ch\u1ecbu \u0111\u01b0\u1ee3c nhi\u1ec7t \u0111\u1ed9 l\u1edbn nh\u1ea5t \u0111\u1ec3 kh\u00f4ng b\u1ecb u\u1ed1n cong l\u00e0: tmax<\/sub> = t = 45\u00baC<\/p>\n\n\n\n B\u00e0i 9 (trang 197 SGK V\u1eadt L\u00fd 10) :<\/strong> X\u00e9t m\u1ed9t\n v\u1eadt r\u1eafn \u0111\u1ed3ng ch\u1ea5t, \u0111\u1eb3ng h\u01b0\u1edbng v\u00e0 c\u00f3 d\u1ea1ng kh\u1ed1i l\u1eadp ph\u01b0\u01a1ng. H\u00e3y ch\u1ee9ng \nminh \u0111\u1ed9 t\u0103ng th\u1ec3 t\u00edch \u0394V c\u1ee7a v\u1eadt r\u1eafn n\u00e0y khi b\u1ecb nung n\u00f3ng t\u1eeb nhi\u1ec7t \u0111\u1ed9 \n\u0111\u1ea7u to<\/sub> \u0111\u1ebfn nhi\u1ec7t \u0111\u1ed9 t \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi c\u00f4ng th\u1ee9c :<\/p>\n\n\n\n \u0394V = V – Vo<\/sub> = \u03b2Vo<\/sub>\u0394t<\/p>\n\n\n\n V\u1edbi Vo<\/sub> v\u00e0 V l\u1ea7n l\u01b0\u1ee3t l\u00e0 th\u1ec3 t\u00edch c\u1ee7a v\u1eadt r\u1eafn \u1edf nhi\u1ec7t \u0111\u1ed9 \u0111\u1ea7u to<\/sub> v\u00e0 nhi\u1ec7t \u0111\u1ed9 cu\u1ed1i t, \u0394t = t – to<\/sub>, \u03b2 \t\u2248 3\u03b1 (\u03b1 l\u00e0 h\u1ec7 s\u1ed1 n\u1edf d\u00e0i c\u1ee7a v\u1eadt r\u1eafn n\u00e0y).<\/p>\n\n\n\n Ch\u00fa \u00fd: \u03b12<\/sup> v\u00e0 \u03b13<\/sup> r\u1ea5t nh\u1ecf so v\u1edbi \u03b1.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n \u1ede nhi\u1ec7t \u0111\u1ed9 t0<\/sub> (\u00baC) c\u1ea1nh h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 l0<\/sub> <\/p>\n\n\n\n \u2192 th\u1ec3 t\u00edch kh\u1ed1i l\u1eadp ph\u01b0\u01a1ng l\u00e0: <\/p>\n\n\n\n \u1ede nhi\u1ec7t \u0111\u1ed9 t (\u00baC) c\u1ea1nh h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 l<\/p>\n\n\n\n \u2192 th\u1ec3 t\u00edch kh\u1ed1i l\u1eadp ph\u01b0\u01a1ng l\u00e0: V = l3<\/sup><\/p>\n\n\n\n M\u1eb7t kh\u00e1c ta c\u00f3: l = l0<\/sub>.(1 + \u03b1\u0394t) \u21d2 V = l0<\/sub>3<\/sup>.(1 + \u03b1\u0394t)3<\/sup> <\/p>\n\n\n\n <\/figure>\n\n\n\n Do \u03b1 r\u1ea5t nh\u1ecf n\u00ean \u03b12<\/sub> v\u00e0 \u03b13<\/sub> c\u0169ng r\u1ea5t nh\u1ecf, ta c\u00f3 th\u1ec3 b\u1ecf qua. <\/p>\n\n\n\n <\/figure>\n\n\n\n \u2192 \u0394V = V \u2013 V0<\/sub> = V0<\/sub>.\u03b2.\u0394t<\/p>\n","protected":false},"excerpt":{"rendered":" B\u00e0i 36 : S\u1ef1 n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a v\u1eadt r\u1eafn C1. ( trang 188 sgk V\u1eadt L\u00fd 10): T\u00ednh h\u1ec7 s\u1ed1 c\u1ee7a m\u1ed7i l\u1ea7n \u0111o ghi trong b\u1ea3ng 36.1. X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a h\u1ec7 s\u1ed1 \u03b1. V\u1edbi sai s\u1ed1 kho\u1ea3ng 5%, nh\u1eadn x\u00e9t xem h\u1ec7 s\u1ed1 \u03b1 c\u00f3 gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i […]<\/p>\n","protected":false},"author":12,"featured_media":44197,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1640],"tags":[],"yoast_head":"\n