C\u00e2u 1:<\/b>\u00a0Trong laze rubi c\u00f3 s\u1ef1 bi\u1ebfn \u0111\u1ed5i n\u0103ng l\u01b0\u1ee3ng t\u1eeb d\u1ea1ng n\u00e0o sang quang n\u0103ng?<\/p>\n

A. Quang n\u0103ng \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. Hi\u1ec7n t\u01b0\u1ee3ng c\u1ea3m \u1ee9ng \u0111i\u1ec7n t\u1eeb<\/p>\n

C. Nhi\u1ec7t n\u0103ng \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. \u0110i\u1ec7n n\u0103ng<\/p>\n

C\u00e2u 2:<\/b>\u00a0Tia laze kh\u00f4ng c\u00f3 \u0111\u1eb7c \u0111i\u1ec3m<\/p>\n

A. \u0111\u1ed9 \u0111\u1ecbnh h\u01b0\u1edbng cao<\/p>\n

B. \u0111\u1ed9 \u0111\u01a1n s\u1eafc cao<\/p>\n

C. c\u01b0\u1eddng \u0111\u1ed9 l\u1edbn<\/p>\n

D. c\u00f4ng su\u1ea5t trung b\u00ecnh c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn<\/p>\n

C\u00e2u 3:<\/b>\u00a0M\u00e0u do m\u1ed9t laze ph\u00e1t ra<\/p>\n

A. m\u00e0u tr\u1eafng<\/p>\n

B. h\u1ed7n h\u1ee3p hai m\u00e0u \u0111\u01a1n s\u1eafc<\/p>\n

C. h\u1ed7n h\u1ee3p nhi\u1ec1u m\u00e0u \u0111\u01a1n s\u1eafc<\/p>\n

D. m\u00e0u \u0111\u01a1n s\u1eafc<\/p>\n

C\u00e2u 4:<\/b>\u00a0T\u00ecm ph\u00e1t bi\u1ec3u sai v\u1ec1 tia laze<\/p>\n

A. tia laze c\u00f3 t\u00ednh \u0111\u1ecbnh h\u01b0\u1edbng cao<\/p>\n

B. tia laze b\u1ecb t\u00e1n s\u1eafc khi qua l\u0103ng k\u00ednh<\/p>\n

C. tia laze l\u00e0 ch\u00f9m s\u00e1ng k\u1ebft h\u1ee3p<\/p>\n

D. tia laze c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 l\u1edbn<\/p>\n

C\u00e2u 5:<\/b>\u00a0Hi\u1ec7u su\u1ea5t c\u1ee7a m\u1ed9t laze<\/p>\n

A l\u1edbn h\u01a1n 100% \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. nh\u1ecf h\u01a1n 100%<\/p>\n

C. b\u1eb1ng 100% \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. r\u1ea5t l\u01a1n so v\u1edbi 100%<\/p>\n

C\u00e2u 6:<\/b>\u00a0T\u00ecm ph\u00e1t bi\u1ec3u sai. C\u00e1c lo\u1ea1i laze th\u00f4ng th\u01b0\u1eddng \u0111\u00e3 \u0111\u01b0\u1ee3c s\u1ea3n xu\u1ea5t l\u00e0<\/p>\n

A. laze r\u1eafn \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. laze kh\u00ed<\/p>\n

C. laze l\u1ecfng \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. laze b\u00e1n d\u1eabnv<\/p>\n

C\u00e2u 7:<\/b>\u00a0Laze kh\u00f4ng \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng<\/p>\n

A. l\u00e0m dao m\u1ed5 trong y h\u1ecdc<\/p>\n

B. x\u00e1c \u0111\u1ecbnh tu\u1ed5i c\u1ed5 v\u1eadt trong ng\u00e0nh kh\u1ea3o c\u1ed5 h\u1ecdc<\/p>\n

C. \u0111\u1ec3 truy\u1ec1n tin b\u1eb1ng c\u00e1p quang<\/p>\n

D. \u0111o c\u00e1c kho\u1ea3ng c\u00e1ch trong ng\u00e0nh tr\u1eafc \u0111\u1ecba<\/p>\n

C\u00e2u 8:<\/b>\u00a0Ng\u01b0\u1eddi ta d\u00f9ng m\u1ed9t laze CO_{2<\/sub>\u00a0c\u00f3 c\u00f4ng su\u1ea5t 8 W \u0111\u1ec3 l\u00e0m dao m\u1ed5. Tia laze chi\u1ebfu v\u00e0o ch\u1ed7 n\u00e0o s\u1ebd l\u00e0m cho n\u01b0\u1edbc c\u1ee7a ph\u1ea7n m\u00f4 \u1edf ch\u1ed7 dod b\u1ed1c h\u01a1i v\u00e0 m\u00f4 b\u1ecb c\u1eaft. Bi\u1ebft nhi\u1ec7t dung ri\u00eang, kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang v\u00e0 nhi\u1ec7t h\u00f3a h\u01a1i c\u1ee7a n\u01b0\u1edbc l\u00e0: c= 4,18 kJ\/kg.K, \u03c1=10^{3<\/sup>\u00a0kg\/m3<\/sup>, L = 2260 kJ\/kg, nhi\u1ec7t \u0111\u1ed9 ban \u0111\u1ea7u c\u1ee7a n\u01b0\u1edbc l\u00e0 37o<\/sup>C. Th\u1ec3 t\u00edch n\u01b0\u1edbc m\u00e0 tia laze l\u00e0m b\u1ed1c h\u01a1i trong 1 s l\u00e0<\/p>\n}}

A. 2,3 mm^{3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 3,9 mm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 3,1 mm3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. 1,6 mm3<\/sup><\/p>\n}

C\u00e2u 9:<\/b>\u00a0Ng\u01b0\u1eddi ta d\u00f9ng m\u1ed9t laze ho\u1ea1t \u0111\u1ed9ng d\u01b0\u1edbi ch\u1ebf \u0111\u1ed9 li\u00ean t\u1ee5c \u0111\u1ec3 khoan m\u1ed9t t\u1ea5m th\u00e9p. C\u00f4ng su\u1ea5t c\u1ee7a ch\u00f9m laze l\u00e0 P = 10 W. \u0110\u01b0\u1eddng k\u00ednh c\u1ee7a ch\u00f9m s\u00e1ng l\u00e0 d = 1 mm, b\u1ec1 d\u00e0y c\u1ee7a t\u1ea5m thep h = 1 mm. Nhi\u1ec7t \u0111\u1ed9 ban \u0111\u1ea7u l\u00e0 t_{1<\/sub>=30^{o<\/sup>C. Bi\u1ebft: Kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a th\u00e9p , \u03c1=7800 kg\/m3<\/sup>; nhi\u1ec7t dung ri\u00eang c\u1ee7a th\u00e9p l\u00e0 c = 448 J\/kg.K ; nhi\u1ec7t n\u00f3ng ch\u1ea3y ri\u00eang c\u1ee7a th\u00e9p \u03bb = 270 kJ\/kg ; \u0111i\u1ec3m n\u00f3ng ch\u1ea3y c\u1ee7a th\u00e9p t2<\/sub>=1535o<\/sup>C. Th\u1eddi gian khoan th\u00e9p l\u00e0<\/p>\n}}

A. 2,3 s \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 0,58 s \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 1,2 s \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. 0,42 s<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0111\u00e1p \u00e1n<\/b><\/p>\n\n\n\n\n

C\u00e2u<\/td>\n	1<\/td>\n	2<\/td>\n	3<\/td>\n	4<\/td>\n	5<\/td>\n	6<\/td>\n	7<\/td>\n	8<\/td>\n	9<\/td>\n<\/tr>\n
\u0110\u00e1p \u00e1n<\/td>\n	A<\/td>\n	D<\/td>\n	D<\/td>\n	B<\/td>\n	B<\/td>\n	C<\/td>\n	B<\/td>\n	C<\/td>\n	B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 8:<\/b>\u00a0C<\/p>\n G\u1ecdi m l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng n\u01b0\u1edbc \u0111\u00e3 b\u1ed1c h\u01a1i th\u00ec Pt=m(c\u0394t\u00b0+L)<\/p>\n <\/p>\n C\u00e2u 9:<\/b>\u00a0B<\/p>\n B\u1ecf qua nhi\u1ec7t l\u01b0\u1ee3ng l\u00e0m n\u00f3ng th\u00e9p v\u00e0 m\u00f4i tr\u01b0\u1eddng xung quanh. Nhi\u1ec7t l\u01b0\u1ee3ng c\u1ea7n truy\u1ec1n l\u00e0 :<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Trong laze rubi c\u00f3 s\u1ef1 bi\u1ebfn \u0111\u1ed5i n\u0103ng l\u01b0\u1ee3ng t\u1eeb d\u1ea1ng n\u00e0o sang quang n\u0103ng? A. Quang n\u0103ng \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. Hi\u1ec7n t\u01b0\u1ee3ng c\u1ea3m \u1ee9ng \u0111i\u1ec7n t\u1eeb C. Nhi\u1ec7t n\u0103ng \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. \u0110i\u1ec7n n\u0103ng C\u00e2u 2:\u00a0Tia laze kh\u00f4ng c\u00f3 \u0111\u1eb7c \u0111i\u1ec3m A. \u0111\u1ed9 \u0111\u1ecbnh h\u01b0\u1edbng cao B. \u0111\u1ed9 \u0111\u01a1n s\u1eafc cao C. c\u01b0\u1eddng \u0111\u1ed9 […]<\/p>\n","protected":false},"author":3,"featured_media":25785,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[158],"tags":[1443,1442],"yoast_head":"\n