Tham kh\u1ea3o \u0111\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n V\u1eadt L\u00fd \u0111\u1ec1 s\u1ed1 7 c\u00f3 \u0111\u00e1p \u00e1n v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft gi\u00fap c\u00e1c em th\u1eed s\u1ee9c v\u1edbi k\u1ef3 thi s\u1eafp t\u1edbi.<\/strong><\/h2>\n

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Xem th\u00eam: \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd<\/p>\n<\/div>\n

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\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n V\u1eadt L\u00fd n\u0103m 2016 – \u0110\u1ec1 s\u1ed1 7<\/strong><\/strong><\/span><\/p>\n

C\u00e2u 1 :  (ID : 83943) <\/strong>trong m\u1ea1ch dao \u0111\u1ed9ng LC c\u00f3 t\u1ee5 \u0111i\u1ec7n l\u00e0 5 \u03bcF, c\u01b0\u1eddng \u0111\u1ed9 t\u1ee9c th\u1eddi c\u1ee7a d\u00f2ng \u0111i\u1ec7n l\u00e0  i = 0,05sin200\u03c0t (A). Bi\u1ec3u th\u1ee9c \u0111i\u1ec7n t\u00edch tr\u00ean t\u1ee5 l\u00e0 <\/p>\n<\/div>\n

A .<\/strong>q = 2,5.10<\/span><\/p>\n

–<\/dd>\n

4\/ \u03c0 cos(200\u03c0t ) (C<\/span>)                  <\/strong><\/p>\n

B.<\/strong>q = 2,5.10<\/p>\n

-5<\/dd>\n

sin(200\u03c0t -\u03c0 \/4) (C)<\/p>\n

C .<\/strong>q = 2,5.10<\/p>\n

-5<\/dd>\n

sin(200\u03c0t \u2013\u03c0\/2 ) (C)      <\/p>\n

D .<\/strong>q = 2,5.10<\/p>\n

-5<\/dd>\n

cos(200\u03c0t \u2013\u03c0\/4 ) (C)<\/p>\n

<\/p>\n

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C\u00e2u 3 :<\/strong>(ID : 84333) <\/strong>\u0111o\u1ea1n m\u1ea1ch MN theo th\u1ee9 t\u1ef1 g\u1ed3m \u0111i\u1ec7n tr\u1edf R, t\u1ee5 \u0111i\u1ec7n C v\u00e0 cu\u1ed9n d\u00e2y kh\u00f4ng thu\u1ea7n c\u1ea3m L, r m\u1eafc n\u1ed1i ti\u1ebfp .A l\u00e0 \u0111i\u1ec3m gi\u1eef t\u1ee5 v\u00e0 cu\u1ed9n d\u00e2y . \u0110\u1eb7t v\u00e0o 2 \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf xoay chi\u1ec1u th\u00ec hi\u1ec7u \u0111i\u1ec7n th\u1ebf U<\/p>\n

MA<\/dd>\n

kh\u00e1c pha \u03c0\/2 so v\u1edbi U<\/p>\n

AN<\/dd>\n

v\u00e0 \u03c0\/6 so v\u1edbi d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch . ph\u01b0\u01a1ng \u00e1n \u0111\u00fang l\u00e0 :<\/p>\n<\/div>\n

A .<\/strong>Kh\u00f4ng \u0111\u1ee7 d\u1eef ki\u1ec7n<\/span><\/p>\n

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B<\/strong>. R=r    <\/p>\n

C<\/strong>.  Z<\/p>\n

L<\/dd>\n

=2Z<\/p>\n

C<\/dd>\n

     <\/p>\n

D<\/strong>.R<r<\/p>\n<\/div>\n

C\u00e2u 4 :<\/strong>(ID : 84334) <\/strong>c\u00f3 2 con l\u1eafc l\u00f2 xo gi\u1ed1ng h\u1ec7t nhau dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a tr\u00ean m\u1eb7t ph\u1eb3ng n\u1eb1m ngang d\u1ecdc theo 2 \u0111\u01b0\u1eddng th\u1eb3ng song song c\u1ea1nh nhau v\u00e0 song song v\u1edbi tr\u1ee5c Ox. Bi\u00ean \u0111\u1ed9 c\u1ee7a con l\u1eafc 1 l\u00e0 A1=3cm , c\u1ee7a con l\u1eafc 2 l\u00e0 A2=6cm . trong qu\u00e1 tr\u00ecnh dao \u0111\u1ed9ng kho\u1ea3ng c\u00e1ch l\u1edbn nh\u1ea5t gi\u1eefa hai v\u1eadt theo ph\u01b0\u01a1ng Ox l\u00e0 a = 3 \u221a3 cm . khi \u0111\u1ed9ng n\u0103ng c\u1ea3 con l\u1eafc 1 l\u00e0 c\u1ef1c \u0111\u1ea1i b\u1eb1ng W th\u00ec \u0111\u1ed9ng n\u0103ng c\u1ee7a con l\u1eafc 2 l\u00e0 :<\/span><\/p>\n

A<\/strong>.2W   <\/span><\/p>\n

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B<\/strong>.W\/2    <\/p>\n

C.<\/strong>2W\/3    <\/p>\n

D.3\/2<\/strong>W<\/p>\n<\/div>\n

C\u00e2u 5 :<\/strong>(ID : 84335) <\/strong> g\u1ecdi v<\/span><\/p>\n

r<\/dd>\n

,v<\/span><\/p>\n

l<\/dd>\n

,v<\/span><\/p>\n

k<\/dd>\n

l\u1ea7n l\u01b0\u1ee3t l\u00e0 t\u1ed1c \u0111\u1ed9 truy\u1ec7n \u00e2m trong ch\u1ea5t r\u1eafn , ch\u1ea5t l\u1ecfng , ch\u1ea5t kh\u00ed . Th\u1ee9 t\u1ef1 s\u1eafp x\u1ebfp \u0111\u00fang l\u00e0  :<\/span><\/p>\n

A.<\/strong>v<\/span><\/p>\n

l<\/dd>\n

>v<\/span><\/p>\n

r<\/dd>\n

>v<\/span><\/p>\n

k<\/dd>\n<\/p>\n

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B.<\/strong>v<\/p>\n

r<\/dd>\n

> v<\/p>\n

l<\/dd>\n

>v<\/p>\n

k<\/dd>\n

    <\/strong><\/p>\n

C.<\/strong>v<\/p>\n

r<\/dd>\n

< v<\/p>\n

l<\/dd>\n

< v<\/p>\n

k<\/dd>\n

        <\/p>\n

D.<\/strong>v<\/p>\n

k<\/dd>\n

> v<\/p>\n

r<\/dd>\n

> v<\/p>\n

l<\/dd>\n<\/p>\n<\/div>\n

C\u00e2u 6 :<\/strong>(ID : 84336) <\/strong> khi \u1edf d\u01b0\u1edbi m\u1eb7t \u0111\u1ea5t v\u00e0 \u1edf c\u00f9ng nhi\u1ec7t \u0111\u1ed9 , m\u1ed9t con l\u1eafc l\u00f2 xo v\u00e0 m\u1ed9t con l\u1eafc \u0111\u01a1n dao \u0111\u1ed9ng v\u1edbi chu k\u00ec b\u1eb1ng nhau T = 2s  .\u0110\u01b0a c\u1ea3 hai con l\u1eafc l\u00ean \u0111\u1ec9nh n\u00fai v\u00e0 gi\u1eef cho nhi\u1ec7t \u0111\u1ed9 kh\u00f4ng \u0111\u1ed5i th\u00ec hai con l\u1eafc dao \u0111\u1ed9ng v\u1edbi chu k\u00ec l\u1ec7ch nhau ch\u00fat \u00edt .Th\u1ec9nh tho\u1ea3ng ch\u00fang l\u1ea1i c\u00f9ng \u0111i qua v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng v\u00e0 chuy\u00ean \u0111\u1ed9ng v\u1ec1 c\u00f9ng m\u1ed9t ph\u00eda , th\u1eddi gian gi\u1eefa hai l\u1ea7n lien ti\u1ebfp nh\u01b0 v\u1eady l\u00e0 8 ph\u00fat 22 gi\u00e2y .chu k\u1ef3 con l\u1eafc \u0111\u01a1n khi \u0111\u00f3 l\u00e0<\/span><\/p>\n

A .<\/strong>2,008s     <\/span><\/p>\n

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B.<\/strong>1,992s     <\/p>\n

C.<\/strong>1,996s     <\/p>\n

D.<\/strong>2,004s<\/p>\n<\/div>\n

C\u00e2u 7 :<\/strong>(ID : 84337)<\/strong>con l\u1eafc \u0111\u01a1n \u0111ang dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi bi\u00ean \u0111\u1ed9 g\u00f3c nh\u1ecf th\u00ec va ch\u1ea1m v\u1edbi m\u1ed9t v\u1eadt nh\u1ecf \u0111ang \u0111\u1ee9ng y\u00ean t\u1ea1i v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng . X\u00e9t 2 tr\u01b0\u1eddng h\u1ee3p  :  1 l\u00e0 va ch\u1ea1m h\u00e0o to\u00e0n \u0111\u00e0n h\u1ed3i ; 2 l\u00e0 va ch\u1ea1m ho\u00e0n to\u00e0n m\u1ec1m. sau khi va ch\u1ea1m , bi\u00ean \u0111\u1ed9 g\u00f3c c\u1ee7a dao \u0111\u1ed9ng<\/span><\/p>\n

A<\/strong>. gi\u1ea3m trong c\u1ea3 2 tr\u01b0\u1eddng h\u1ee3p<\/span><\/p>\n

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B.<\/strong>t\u0103ng trong c\u1ea3 2 tr\u01b0\u1eddng h\u1ee3p<\/p>\n

C .<\/strong>c\u00f3 th\u1ec3 t\u0103ng khi va ch\u1ea1m \u0111\u00e0n h\u1ed3i<\/p>\n

D.<\/strong>ch\u1ec9 gi\u1ea3m khi va ch\u1ea1m m\u1ec1m<\/p>\n<\/div>\n

C\u00e2u 8 : (ID : 84338)<\/strong>D\u00f9ng kh\u00e1ng c\u1ea3u 1 \u0111o\u1ea1n mafch PLC n\u1ed1i ti\u1ebfp c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn h\u01a1n gi\u00e1 trj c\u1ee7a c\u1ea3m kh\u00e1ng . C\u00f3 th\u1ec3 l\u00e0m cho hi\u1ec7n t\u01b0\u1ee3ng c\u1ed9ng h\u01b0\u1edfng \u0111i\u1ec7n x\u1ea3y ra b\u1eb1ng c\u00e1ch<\/span><\/p>\n

A.<\/strong>t\u0103ng h\u1ec7 s\u00f4 t\u1ef1 c\u1ea3m c\u1ee7a cu\u1ed9n d\u00e2y<\/span><\/p>\n

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B.<\/strong>  t\u0103ng \u0111i\u1ec7n dung c\u1ee7a t\u1ee5 \u0111i\u1ec7n<\/p>\n

C.<\/strong>gi\u1ea3m \u0111i\u1ec7n tr\u1edf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch<\/p>\n

D.<\/strong>gi\u1ea3m t\u1ea7n s\u1ed1 d\u00f2ng \u0111i\u1ec7n<\/p>\n<\/div>\n

C\u00e2u 9 : (ID : 84339)<\/strong>M\u1ea1ch \u0111i\u1ec7n xoay chi\u1ec1u g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi cu\u1ed9n d\u00e2y. \u0110\u1eb7t v\u00e0o hai \u0111\u1ea7u m\u1ea1ch \u0111i\u1ec7n m\u1ed9t \u0111i\u1ec7n \u00e1p xoay chi\u1ec1u u = 100\u221a6 sin(100 \u03c0t) (V). D\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch l\u1ec7ch pha \u03c0\/6 so v\u1edbi u v\u00e0 l\u1ec7ch pha \u03c0\/3 so v\u1edbi u<\/span><\/p>\n

d   .<\/dd>\n

\u0110i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng \u1edf hai \u0111\u1ea7u cu\u1ed9n d\u00e2y c\u00f3 gi\u00e1 tr\u1ecb.<\/span><\/p>\n

A.<\/strong>100(V)                                            <\/span><\/p>\n

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B.<\/strong>100\u221a  3(V)<\/p>\n

C.<\/strong>200(V)<\/p>\n

D.<\/strong>100\u221a2(V)<\/p>\n<\/div>\n

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C\u00e2u 10: (ID : 84340)<\/strong>Ta thu \u0111\u01b0\u1ee3c quang ph\u1ed5 v\u1ea1ch ph\u00e1t x\u1ea1 khi<\/p>\n<\/div>\n

A.<\/strong>\u0111un n\u01b0\u1edbc s\u00f4i \u1edf nhi\u1ec7t \u0111\u1ed9 \u0111\u1ee7 cao.<\/span><\/p>\n

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B.<\/strong>nung m\u1ed9t c\u1ee5c s\u1eaft t\u1edbi nhi\u1ec7t \u0111\u1ed9 \u0111\u1ee7 cao.<\/p>\n

C.<\/strong>nung n\u00f3ng h\u01a1i th\u1ee7y ng\u00e2n cao \u00e1p.<\/p>\n

D.<\/strong>cho tia l\u1eeda \u0111i\u1ec7n ph\u00f3ng qu\u00e1 kh\u00ed Hi\u0111r\u00f4 r\u1ea5t lo\u00e3ng.<\/p>\n<\/div>\n

C\u00e2u 11:  (ID : 84341)<\/strong>M\u1ed9t v\u1eadt th\u1ef1c hi\u1ec7n \u0111\u1ed3ng th\u1eddi hai giao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng t\u1ea7n s\u1ed1 x<\/span><\/p>\n

1 =  <\/dd>\n

5cos(10\u03c0t \u2013 \u03c0\/3)cm v\u00e0 x<\/span><\/p>\n

2<\/dd>\n

= 5sin(10\u03c0t + \u03c0\/2)cm. T\u1ed1c \u0111\u1ed9 trung b\u00ecnh c\u1ee7a v\u1eadt t\u1eeb \u0111\u1ea7u chuy\u1ec3n \u0111\u1ed9ng \u0111\u1ebfn khi qua VTCB l\u1ea7n \u0111\u1ea7u ti\u00ean l\u00e0:<\/span><\/p>\n

A.<\/strong>0,47m\/s<\/span><\/p>\n

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B.<\/strong>1,47 m\/s<\/p>\n

C.<\/strong>2,47 m\/s<\/p>\n

D.<\/strong>0,87 m\/s<\/p>\n<\/div>\n

C\u00e2u 12 :<\/strong>(ID : 84342)<\/strong>Con l\u1eafc l\u00f2 xo g\u1ed3m v\u1eadt n\u1eb7ng c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng 200g v\u00e0 l\u00f2 xo c\u00f3 \u0111\u1ed9 c\u1ee9ng  200N\/m . K\u00e9o v\u1eadt t\u1edbi v\u1ecb tr\u00ed c\u00f3 li \u0111\u1ed9 b\u1eb1ng 2cm r\u1ed3i truy\u1ec1n ho v\u1eadt c\u00f3 v\u1eadn t\u1ed1c 1,095 m\/s theo chi\u1ec1u d\u01b0\u01a1ng c\u1ee7a tr\u1ee5c t\u1ecda \u0111\u1ed9. Ch\u1ecdn g\u1ed1c th\u1eddi gian l\u00e0 th\u1eddi \u0111i\u1ec3m k\u00edch th\u00edch cho v\u1eadt giao \u0111\u1ed9ng. Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c t\u1eeb  th\u1eddi \u0111i\u1ec3m 1\/15s \u0111\u1ebfn th\u1eddi \u0111i\u1ec3m \u00bc s l\u00e0<\/span><\/p>\n

A.<\/strong>14,67cm<\/span><\/p>\n

B.<\/strong>15,46 cm<\/p>\n

C.<\/strong>14,54 cm<\/p>\n

D.<\/strong>14cm<\/p>\n

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C\u00e2u 13 : (ID : 84343)<\/strong>M\u00e1y bi\u1ebfn th\u1ebf c\u00f3 110 v\u00f2ng \u1edf cu\u1ed9n s\u01a1 c\u1ea5p v\u00e0 220 v\u00f2ng \u1edf cu\u1ed9n th\u1ee9 c\u1ea5p. Cu\u1ed9n d\u00e2y s\u01a1 c\u1ea5p c\u00f3 \u0111i\u1ec7n tr\u1edf thu\u1ea7n r = 3 \u2126   v\u00e0 c\u1ea3m kh\u00e1ng Z<\/p>\n

L<\/dd>\n

= 4 \u2126  . N\u1ed1i hai \u0111\u1ea7u cu\u1ed9n s\u01a1 c\u1ea5p v\u1edbi hi\u1ec7u \u0111i\u1ec7n th\u1ebf xoay chi\u1ec1u c\u00f3 40 V th\u00ec hi\u1ec7u \u0111i\u1ec7n th\u1ebf \u1edf hai \u0111\u1ea7u cu\u1ed9n th\u1ee9 c\u1ea5p \u0111\u1ec3 h\u1edf l\u00e0:<\/p>\n<\/div>\n

A.<\/strong>64V<\/span><\/p>\n

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B.<\/strong>80V<\/p>\n

C<\/strong>. 32V<\/p>\n

D.<\/strong>72 V<\/p>\n<\/div>\n

C\u00e2u 14 :  (ID : 84344)<\/strong>M\u1ed9t \u0111\u00e0i b\u00e1n d\u1eabn  c\u00f3 th\u1ec3 thu c\u1ea3 s\u00f3ng  AM v\u00e0 d\u1ea3i s\u00f3ng FM b\u1eb1ng c\u00e1ch thay \u0111\u1ed5i cu\u1ed9n c\u1ea3m L c\u1ee7a m\u1ea1ch thu s\u00f3ng nh\u01b0ng v\u1eabn d\u00f9ng chung m\u1ed9t t\u1ee5 xoay. Khi thu s\u00f3ng AM, \u0111\u00e0i thu \u0111\u01b0\u1ee3c d\u1ea3i s\u00f3ng t\u1eeb 100m \u0111\u1ebfn 600m. Khi thu s\u00f3ng FM, \u0111\u00e0i thu \u0111\u01b0\u1ee3c b\u01b0\u1edbc s\u00f3ng ng\u1eafn nh\u1ea5t l\u00e0 2,5m. B\u01b0\u1edbc s\u00f3ng d\u00e0i nh\u1ea5t trong d\u1ea3i  s\u00f3ng FM m\u00e0 \u0111\u00e0i thu \u0111\u01b0\u1ee3c l\u00e0<\/span><\/p>\n

A.<\/strong>5m<\/span><\/p>\n

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B.<\/strong>7,5 m<\/p>\n

C.<\/strong>15m<\/p>\n

D.<\/strong>12m<\/p>\n<\/div>\n

C\u00e2u 15: (ID : 84345)<\/strong>Trong c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng \u0111\u1eb7c tr\u01b0ng cho d\u00f2ng \u0111i\u1ec7n xoay chi\u1ec1u sau \u0111\u00e2y, gi\u00e1 tr\u1ecb hi\u1ec7u d\u1ee5ng \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0o<\/span>kh\u00f4ng<\/strong>c\u00f3 \u00fd ngh\u0129a th\u1ef1c ti\u1ec5n?<\/span><\/p>\n

A .<\/strong>  \u0110i\u1ec7n \u00e1p<\/span><\/p>\n

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B.<\/strong>   Su\u1ea5t \u0111i\u1ec7n \u0111\u1ed9ng<\/p>\n

C.<\/strong>C\u00f4ng su\u1ea5t<\/p>\n

D.<\/strong>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n<\/p>\n<\/div>\n

C\u00e2u 16:<\/strong>(ID : 84346)<\/strong>\u0110\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m m\u1ed9t cu\u1ed9n d\u00e2y l\u00fd t\u01b0\u1edfng c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L<\/span><\/p>\n

0 \u00ad<\/dd>\n

m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi h\u1ed9p \u0111en X m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf xoay chi\u1ec1u u = U<\/span><\/p>\n

0<\/dd>\n

cos(\u03c9t + \u03c0\/6) (V) th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n t\u1ee9c th\u1eddi ch\u1ea1y trong m\u1ea1ch la i=I<\/span><\/p>\n

0<\/dd>\n

cos(\u03c9t+ \u03c0\/6) (A). H\u1ed9p \u0111en X c\u00f3 th\u1ec3 ch\u1ee9a<\/span><\/p>\n

A.<\/strong>T\u1ee5 \u0111i\u1ec7n<\/span><\/p>\n

B.<\/strong>cu\u1ed9n d\u00e2y thu\u1ea7n c\u1ea3m<\/p>\n

C.<\/strong>\u0111i\u1ec7n tr\u1edf thu\u1ea7n v\u00e0 t\u1ee5 \u0111i\u1ec7n<\/p>\n

D.<\/strong>\u0111i\u1ec7n tr\u1edf v\u00e0 cu\u1ed9n d\u00e2y<\/p>\n

\n

C\u00e2u 17: (ID : 84347)<\/strong>Con l\u1eafc \u0111\u01a1n \u0111ang dao \u0111\u1ed9ng v\u1edbi chu k\u00ec 1s t\u1ea1i n\u01a1i c\u00f3 gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng b\u1eb1ng 10m\/s<\/p>\n

2  <\/dd>\n

. L\u1ea5y \u03c0<\/p>\n

2<\/dd>\n

= 10. V\u1eadt nh\u1ecf c\u1ee7a con l\u1eafc c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng 50g. L\u1ef1c h\u1ed3i ph\u1ee5c c\u1ef1c \u0111\u1ea1i t\u00e1c d\u1ee5ng l\u00ean con l\u1eafc b\u1eb1ng 0,05N. L\u1ef1c c\u0103ng d\u00e2y treo con l\u1eafc khi v\u1eadt nh\u1ecf \u0111i qua v\u1ecb tr\u00ed th\u1ebf n\u0103ng b\u1eb1ng m\u1ed9t n\u1eeda \u0111\u1ed9ng n\u0103ng l\u00e0<\/p>\n<\/div>\n

A.<\/strong>0,5050(N)<\/span><\/p>\n

\n

B.<\/strong>0,4975(N)<\/p>\n

C.(<\/strong>0,4950(N)<\/p>\n

D.<\/strong>0,5025(N).<\/p>\n<\/div>\n

C\u00e2u 18: (ID : 84348)<\/strong>Trong th\u00ed nghi\u1ec7m giao thoa v\u1edbi 2 khe Young, ngu\u1ed3n ph\u00e1t \u00e1nh s\u00e1ng c\u00f3 b\u01b0\u1edbc s\u00f3ng 0,42 \u03bcm \u2264 \u03bb \u2264 0,6 \u03bcm. Kho\u1ea3ng c\u00e1ch gi\u1eefa hai khe l\u00e0 0,15 mm, kho\u1ea3ng c\u00e1ch t\u1eeb 2 khe t\u1edbi m\u00e0n l\u00e0 60cm. Kho\u1ea3ng c\u00e1ch nh\u1ecf nh\u1ea5t t\u1eeb v\u1eadt trung t\u00e2m t\u1edbi v\u1ecb tr\u00ed c\u00f3 \u0111\u1ed3ng th\u1eddi c\u1ef1c \u0111\u1ea1i giao thoa c\u1ee7a 2 b\u1ee9c x\u1ea1 l\u00e0<\/span><\/p>\n

A<\/strong>.6,72mm<\/span><\/p>\n

\n

B.<\/strong>7,2mm<\/p>\n

C.<\/strong>5,04mm<\/p>\n

D.<\/strong>4,8mm<\/p>\n<\/div>\n

C\u00e2u 19: (ID : 84349)<\/strong>M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00e0o d\u1ecdc theo  tr\u1ee5c Ox, g\u1ecdi \u2206t l\u00e0 kho\u1ea3ng th\u1eddi gian gi\u1eefa hai l\u1ea7n li\u00ean ti\u1ebfp c\u00f3 \u0111\u1ed9ng n\u0103ng b\u1eb1ng th\u1ebf n\u0103ng. T\u1ea1i th\u1eddi \u0111i\u1ec3m t v\u1eadt qua v\u1ecb tr\u00ed c\u00f3 t\u1ed1c \u0111\u1ed9 8 \u03c0 \u221a3 cm\/s v\u1edbi \u0111\u1ed9 l\u1edbn gia t\u1ed1c 96 \u03c0<\/span><\/p>\n

2<\/dd>\n

cm\/s<\/span><\/p>\n

2<\/dd>\n

, sau \u0111\u00f3 m\u1ed9t kho\u1ea3ng th\u1eddi gian \u0111\u00fang b\u1eb1ng \u2206t v\u1eadt qua v\u1ecb tr\u00ed c\u00f3 \u0111\u1ed9 l\u1edbn v\u1eadn t\u1ed1c 24 \u03c0cm\/s. Bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng c\u1ee7a v\u1eadt l\u00e0<\/span><\/p>\n

A<\/strong>.4\u221a2 cm<\/span><\/p>\n

\n

B.<\/strong>8cm<\/p>\n

C.<\/strong>4\u221a3 cm<\/p>\n

D.<\/strong>5\u221a2cm<\/p>\n<\/div>\n

C\u00e2u 20: (ID : 84350)<\/strong>M\u1ea1ch dao \u0111\u1ed9ng LC l\u00ed t\u01b0\u1edfng c\u00f3 L= 5\u03bcH v\u00e0 C = 8nF. T\u1ea1i th\u1eddi \u0111i\u1ec3m t,  t\u1ee5 \u0111i\u1ec7n ph\u00f3ng \u0111i\u1ec7n v\u00e0 \u0111i\u1ec7n t\u00edch c\u1ee7a t\u1ee5 t\u1ea1i th\u1eddi \u0111i\u1ec3m \u0111\u00f3 q= 2,4.10<\/span><\/p>\n

-8<\/dd>\n

C. T\u1ea1i th\u1eddi \u0111i\u1ec3m sau \u0111\u00f3 \u2206t = \u03c0(\u03bcs) th\u00ec hi\u1ec7u \u0111i\u1ec7n th\u1ebf tr\u00ean t\u1ee5 l\u00e0<\/span><\/p>\n

A.<\/strong>3(V)<\/span><\/p>\n

B.-<\/strong>3(V)<\/p>\n

C.<\/strong>3,6(V)<\/p>\n

D.-<\/strong>4,8(V)<\/p>\n

\n

C\u00e2u 21: (ID : 84351)<\/strong>M\u1ed9t khung d\u00e2y c\u00f3 \u0111i\u1ec7n t\u00edch 400cm<\/p>\n

2<\/dd>\n

g\u1ed3m 100 v\u00f2ng, quay \u0111\u1ec1u trong t\u1eeb tr\u01b0\u1eddng \u0111\u1ec1u B = 0,15T quanh m\u1ed9t tr\u1ee5c n\u1eb1m trong m\u1ed9t m\u1eb7t ph\u1eb3ng khung v\u1edbi t\u1ed1c \u0111\u1ed9 3600 v\u00f2ng\/ph\u00fat. C\u00e1c \u0111\u01b0\u1eddng c\u1ea3m \u1ee9ng t\u1eeb lu\u00f4n lu\u00f4n vu\u00f4ng g\u00f3c v\u1edbi tr\u1ee5c quay. N\u1ebfu<\/p>\n

t = 0 , vect\u01a1 c\u1ea3m \u1ee9ng t\u1eebB<\/strong>ng\u01b0\u1ee3c h\u01b0\u1edbng v\u1edbi ph\u00e1p tuy\u1ebfn c\u1ee7a m\u1eb7t khung th\u00ec l\u00fac t = 1\/720 s, su\u1ea5t \u0111i\u1ec7n \u0111\u1ed9ng trong khung c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng<\/p>\n<\/div>\n

A.-<\/strong>127(V)<\/span><\/p>\n

\n

B.<\/strong>127(V)<\/p>\n

C.-<\/strong>113(V)<\/p>\n

D.<\/strong>113(V)<\/p>\n<\/div>\n

C\u00e2u 22:<\/strong>(ID : 84352)<\/strong>M\u1ed9t m\u1ea1ch RLC n\u1ed1i ti\u1ebfp c\u00f3 R thay \u0111\u1ed5i \u0111\u01b0\u1ee3c . Khi cho R = R<\/span><\/p>\n

1<\/dd>\n

= 10 \u2126 ho\u1eb7c<\/span><\/p>\n

\n

R = R<\/p>\n

2<\/dd>\n

= 30 \u2126 th\u00ec c\u00f4ng c\u1ee7a m\u1ea1ch nh\u01b0 nhau. H\u1ec7 s\u1ed1 c\u00f4ng su\u1ea5t c\u1ee7a m\u1ea1ch khi R = R<\/p>\n

1 <\/dd>\n

b\u1eb1ng<\/p>\n<\/div>\n

A<\/strong>.\u221a2\/2<\/span><\/p>\n

\n

B<\/strong>. 1<\/p>\n

C.<\/strong>\u221a3\/2<\/p>\n

D.<\/strong>1\/2<\/p>\n<\/div>\n

C\u00e2u 23: (ID : 84353)<\/strong>Th\u00ed nghi\u1ec7m Young giao thoa \u00e1nh s\u00e1ng: ngu\u1ed3n s\u00e1ng ph\u00e1t \u0111\u1ed3ng th\u1eddi hai b\u1ee9c x\u1ea1 c\u00f3 b\u01b0\u1edbc s\u00f3ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u03bb<\/span><\/p>\n

1<\/dd>\n

= 0,96\u03bcm  v\u00e0 l\u00e0 \u03bb<\/span><\/p>\n

2<\/dd>\n

= 0,46 \u03bcm,  2 \u0111i\u1ec3m P,Q \u1edf hai ph\u00eda \u0111\u1ed1i v\u1edbi v\u00e2n s\u00e1ng trung t\u00e2m O: t\u1ea1i P l\u00e0 v\u00e2n s\u00e1ng b\u1eadc 6 \u1ee9ng v\u1edbi b\u01b0\u1edbc s\u00f3ng \u03bb<\/span><\/p>\n

1 <\/dd>\n

v\u00e0 t\u1ea1i \u03bb<\/span><\/p>\n

2<\/dd>\n

. Tr\u00ean \u0111o\u1ea1n PQ ta \u0111\u1ebfm \u0111\u01b0\u1ee3c<\/span><\/p>\n

A.<\/strong>21 v\u00e2n s\u00e1ng<\/span><\/p>\n

\n

B.<\/strong>19 v\u00e2n s\u00e1ng<\/p>\n

C.<\/strong>27 v\u00e2n s\u00e1ng<\/p>\n

D.<\/strong>20 v\u00e2n s\u00e1ng<\/p>\n<\/div>\n

C\u00e2u 24 :<\/strong>(ID : 84354)<\/strong>Ngu\u1ed3n<\/span>kh\u00f4ng<\/strong>ph\u00e1t tia t\u1eed ngo\u1ea1i l\u00e0<\/span><\/p>\n

A.<\/strong>\u0111\u00e8n h\u01a1i th\u1ee7y ng\u00e2n<\/span><\/p>\n

\n

B.<\/strong>M\u1eb7t tr\u1eddi<\/p>\n

C.<\/strong>\u0111\u00e8n h\u01a1i natri<\/p>\n

D<\/strong>.h\u1ed3 quang \u0111i\u1ec7n<\/p>\n<\/div>\n

C\u00e2u 25: (ID : 84355)<\/strong>ho \u0111o\u1ea1n m\u1ea1ch xoay chi\u1ec1u n\u1ed1i ti\u1ebfp AB theo th\u1ee9 t\u1ef1 g\u1ed3m cu\u1ed9n d\u00e2y thu\u1ea7n c\u1ea3m L, \u0111i\u1ec7n tr\u1edf R v\u00e0 t\u1ee5 \u0111i\u1ec7n C. M l\u00e0 \u0111i\u1ec3m n\u1eb1m gi\u1eefa cu\u1ed9n d\u00e2y v\u00e0 \u0111i\u1ec7n tr\u1edf c\u00f2n N l\u00e0 \u0111i\u1ec3m n\u1eb1m gi\u1eefa \u0111i\u1ec7n tr\u1edf v\u00e0 t\u1ee5. \u0110\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch AB \u0111i\u1ec7n \u00e1p xoay chi\u1ec1u U<\/span><\/p>\n

AB<\/dd>\n

= U<\/span><\/p>\n

0<\/dd>\n

cos(100 \u03c0t  + \u03c6) th\u00ec \u0111i\u1ec7n \u00e1p tr\u00ean c\u00e1c \u0111o\u1ea1n m\u1ea1ch AN v\u00e0 MB l\u00e0<\/span><\/p>\n

\n

U<\/p>\n

AN<\/dd>\n

= 100\u221a2cos100 \u03c0t  (V)   v\u00e0 U<\/p>\n

MB =<\/dd>\n

100\u221a6cos(100 \u03c0t  – \u03c0\/2)(V). Gi\u00e1 tr\u1ecb U<\/p>\n

0<\/dd>\n

 b\u1eb1ng<\/p>\n<\/div>\n

A<\/strong>.50\u221a14(V)<\/span><\/p>\n

B.<\/strong>25\u221a14(V)<\/p>\n

C.<\/strong>100\u221a14(V)<\/p>\n

D.<\/strong>75\u221a14(V)<\/p>\n

\n

C\u00e2u 26:<\/strong> (ID : 84356)<\/strong>Trong th\u00ed nghi\u1ec7m giao thoa s\u00f3ng tr\u00ean m\u1eb7t n\u01b0\u1edbc, 2 ngu\u1ed3n k\u1ebft h\u1ee3p A,B c\u00e1ch nhau 8cm ph\u00e1t ra 2 s\u00f3ng c\u00f9ng pha. T\u1ea1i \u0111i\u1ec3m M tr\u00ean m\u1eb7t n\u01b0\u1edbc (MA = 25cm, MB 20,5cm), s\u00f3ng c\u00f3 bi\u00ean \u0111\u1ed9 c\u1ef1c \u0111\u1ea1i gi\u1eefa M v\u00e0 trung tr\u1ef1c c\u1ee7a AB c\u00f3 2 d\u00e2y c\u1ef1c \u0111\u1ea1i kh\u00e1c. C v\u00e0 D l\u00e0 2 \u0111i\u1ec3m tr\u00ean m\u1eb7t n\u01b0\u1edbc sao cho ABCD l\u00e0 h\u00ecnh vu\u00f4ng. S\u1ed1 \u0111i\u1ec3m giao \u0111\u1ed9ng c\u00f3 bi\u00ean \u0111\u1ed9 c\u1ef1c ti\u1ec3u tr\u00ean \u0111o\u1ea1n AC b\u1eb1ng<\/p>\n<\/div>\n

A<\/strong>.9<\/span><\/p>\n

\n

B.<\/strong>7<\/p>\n

C.<\/strong>5<\/p>\n

D.<\/strong>3<\/p>\n<\/div>\n

C\u00e2u 27: (ID : 84357)<\/strong>M\u1ed9t t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u00ean dung C = 5,3\u03bcF m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi \u0111i\u1ec7n tr\u1edf R = 300\u2126 th\u00e0nh m\u1ed9t \u0111o\u1ea1n m\u1ea1ch. M\u1eafc \u0111o\u1ea1n m\u1ea1ch n\u00e0y v\u00e0o m\u1ea1ng \u0111i\u1ec7n xoay chi\u1ec1u 220V \u2013 50Hz. \u0110i\u1ec7n n\u0103ng m\u00e0 \u0111o\u1ea1n m\u1ea1ch ti\u00eau th\u1ee5 trong m\u1ed9t ng\u00e0y \u0111\u00eam l\u00e0<\/span><\/p>\n

A.<\/strong>1,97.10<\/span><\/p>\n

6<\/dd>\n

J<\/span><\/p>\n

\n

B.<\/strong>1,0952 kWh<\/p>\n

C.<\/strong>0,5476 kWh<\/p>\n

D.<\/strong>0,7744 kWh<\/p>\n<\/div>\n

C\u00e2u 28: (ID : 84358)<\/strong>Cho m\u1ed9t s\u00f3ng ngang c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh s\u00f3ng l\u00e0: u = 4sin2 \u03c0(t+ x\/-5)mm, trong \u0111\u00f3 x t\u00ednh b\u1eb1ng cm, t t\u00ednh b\u1eb1ng gi\u00e2y. T\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng l\u00e0<\/span><\/p>\n

A.-<\/strong>5cm\/s<\/span><\/p>\n

\n

B.-<\/strong>5m\/s<\/p>\n

C<\/strong>.5cm\/s<\/p>\n

D.<\/strong>5mm\/s<\/p>\n<\/div>\n

C\u00e2u 29: (ID : 84359)<\/strong>Th\u1ef1c hi\u1ec7n giao thoa \u00e1nh s\u00e1ng v\u1edbi khe Young: kho\u1ea3ng c\u00e1ch gi\u1eefa c\u00e1c khe S<\/span><\/p>\n

1<\/dd>\n

S<\/span><\/p>\n

2<\/dd>\n

 c\u00f3 th\u1ec3 thay \u0111\u1ed5i \u0111\u01b0\u1ee3c, kho\u1ea3ng c\u00e1ch t\u1eeb hai khe \u0111\u1ebfn m\u00e0n l\u00e0 v\u00e0 b\u01b0\u1edbc s\u00f3ng \u00e1nh s\u00e1ng chi\u1ebfu v\u00e0o hai khe kh\u00f4ng \u0111\u1ed5i. X\u00e9t hai \u0111i\u1ec3m M v\u00e0 N tr\u00ean m\u00e0n n\u1eb1m \u0111\u1ed1i x\u1ee9ng v\u1ec1 hai ph\u00eda so v\u1edbi v\u00e2n trung t\u00e2m. Khi t\u0103ng kho\u1ea3ng c\u00e1ch gi\u1eefa hai khe S<\/span><\/p>\n

1<\/dd>\n

S<\/span><\/p>\n

2 l\u00ean<\/dd>\n

3 l\u1ea7n th\u00ec s\u1ed1 v\u00e2n s\u00e1ng quan s\u00e1t \u0111\u01b0\u1ee3c tr\u00ean \u0111o\u1ea1n MN<\/span><\/p>\n

A.<\/strong>gi\u1ea3m<\/span><\/p>\n

\n

B.<\/strong>gi\u1ea3m 3 l\u1ea7n<\/p>\n

C.<\/strong>t\u0103ng<\/p>\n

D.<\/strong>t\u0103ng 3 l\u1ea7n<\/p>\n<\/div>\n

C\u00e2u 30: (ID : 84360)<\/strong>M\u1ed9t s\u00f3ng \u0111i\u1ec7n t\u1eeb truy\u1ec1n \u0111i theo h\u01b0\u1edbng \u0110\u00f4ng \u2013 T\u00e2y. Khi  vect\u01a1 t\u1eeb tr\u01b0\u1eddng c\u00f3 \u0111\u1ed9 l\u1edbn b\u1eb1ng n\u1eeda  gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i v\u00e0 c\u00f3 ph\u01b0\u01a1ng Nam \u2013 B\u1eafc th\u00ec vect\u01a1 \u0111i\u1ec7n tr\u01b0\u1eddng<\/span><\/p>\n

\n

A<\/strong>.b\u1eb1ng n\u1eeda gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i v\u00e0 h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng t\u1eeb tr\u00ean xu\u1ed1ng<\/p>\n

B.<\/strong>\u0111\u1ea1t c\u1ef1c \u0111\u1ea1i v\u00e0 h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng t\u1eeb d\u01b0\u1edbi l\u00ean<\/p>\n

C.<\/strong>\u0111\u1ea1t c\u1ef1c \u0111\u1ea1i v\u00e0 h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng t\u1eeb tr\u00ean xu\u1ed1ng<\/p>\n

D.<\/strong>b\u1eb1ng n\u1eeda gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i v\u00e0 h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng t\u1eeb d\u01b0\u1edbi l\u00ean<\/p>\n

C\u00e2u 31: (ID : 84361)<\/strong>M t\u1edbi N l\u00e0 hai \u0111i\u1ec3m tr\u00ean c\u00f9ng ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng n\u00ean m\u1eb7t n\u01b0\u1edbc c\u00e1ch ngu\u1ed3n theo th\u1ee9 t\u1ef1 d<\/p>\n

1<\/dd>\n

= 5cm v\u00e0 d<\/p>\n

2<\/dd>\n

= 20cm. Bi\u1ebft r\u1eb1ng c\u00e1c v\u00f2ng tr\u00f2n \u0111\u1ed3ng t\u00e2m c\u1ee7a s\u00f3ng nh\u1eadn \u0111\u01b0\u1ee3c n\u0103ng l\u01b0\u1ee3ng dao \u0111\u1ed9ng nh\u01b0 nhau. T\u1ea1i M, ph\u01b0\u01a1ng tr\u00ecnh s\u00f3ng c\u00f3 d\u1ea1ng U<\/p>\n

M<\/dd>\n

= 5cos(10 \u03c0t + \u03c0\/3)(cm). V\u1eadn t\u1ed1c truy\u1ec1n s\u00f3ng v = 30cm\/s. T\u1ea1i th\u1eddi \u0111i\u1ec3m t th\u00ec \u0111i\u1ec3m M c\u00f3 li \u0111\u1ed9 U<\/p>\n

M<\/dd>\n

(t) = 4cm th\u00ec N c\u00f3 li \u0111\u1ed9 l\u00e0<\/p>\n<\/div>\n

A.<\/strong>4cm<\/span><\/p>\n

\n

B.<\/strong>2cm<\/p>\n

C<\/strong>.-4cm<\/p>\n

D.-<\/strong>2cm<\/p>\n<\/div>\n

C\u00e2u 32: (ID : 84362)<\/strong>Cho \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n xoay chi\u1ec1u MN theo th\u1ee9 t\u1ef1 g\u1ed3m ampe k\u1ebf l\u00fd t\u01b0\u1edfng, \u0111i\u1ec7n tr\u1edf thu\u1ea7n R , cu\u1ed9n c\u1ea3m thu\u1ea7n R v\u00e0 t\u1ee5 \u0111i\u1ec7n C, P l\u00e0 \u0111i\u1ec3m gi\u1eefa cu\u1ed9n c\u1ea3m v\u00e0 t\u1ee5\u0111i\u1ec7n, gi\u1eefa P v\u00e0 N c\u00f3 g\u1eafn th\u00eam m\u1ed9t kh\u00f3a K c\u00f3 \u0111i\u1ec7n tr\u1edf kh\u00f4ng \u0111\u00e1ng k\u1ec3. Bi\u1ebft U<\/span><\/p>\n

NM<\/dd>\n

= 100cos100 \u03c0t(V). Khi kh\u00f3a K m\u1edf, d\u00f9ng m\u1ed9t v\u00f4n k\u1ebf l\u00ed t\u01b0\u1edfng \u0111o \u0111\u01b0\u1ee3c U<\/span><\/p>\n

MP =<\/dd>\n

U<\/span><\/p>\n

1 \u00ad<\/dd>\n

, U<\/span><\/p>\n

PN<\/dd>\n

=<\/span>U<\/span><\/p>\n

2<\/dd>\n

= 1,2 U<\/span><\/p>\n

1<\/dd>\n

, \u0111\u1ed3ng th\u1eddi ampe k\u1ebf ch\u1ec9 0,5 A. khi kh\u00f3a K \u0111\u00f3ng ampe k\u1ebf ch\u1ec9 0,5A. C\u1ea3m kh\u00e1ng Z<\/span><\/p>\n

L<\/dd>\n

 c\u1ee7a cu\u1ed9n d\u00e2y c\u00f3 gi\u00e1 tr\u1ecb l\u00e0<\/span><\/p>\n

A.<\/strong>120 \u2126<\/span><\/p>\n

\n

B.<\/strong>80 \u2126<\/p>\n

C<\/strong>.160 \u2126<\/p>\n

D.<\/strong>180 \u2126<\/p>\n<\/div>\n

C\u00e2u 33: (ID : 84363)<\/strong>Ch\u1ecdn c\u00e2u \u0111\u00fang<\/span><\/p>\n

\n

A.<\/strong>C\u00e1c d\u1ea1o \u0111\u1ed9ng t\u1eaft d\u1ea7n \u0111\u1ec1u c\u1ea7n \u0111\u01b0\u1ee3c duy tr\u00ec<\/p>\n

B.<\/strong>C\u00e1c dao \u0111\u1ed9ng t\u1eaft d\u1ea7n \u0111\u1ec1u c\u00f3 h\u1ea1i<\/p>\n

.<\/dd>\n<\/p>\n

C.<\/strong>C\u00e1ch \u0111\u01a1n gi\u1ea3n nh\u1ea5t l\u00e0m cho h\u1ec7 dao \u0111\u1ed9ng kh\u00f4ng t\u1eaft l\u00e0 cung c\u1ea5p n\u0103ng l\u01b0\u1ee3ng cho n\u00f3 sau m\u1ed7i chu k\u00ec \u0111\u1ec3 b\u00f9 l\u1ea1i  n\u0103ng l\u01b0\u1ee3ng \u0111\u00e3 m\u1ea5t.<\/p>\n

D.<\/strong>Dao \u0111\u1ed9ng c\u1ee7a gi\u1ea3m x\u00f3c xe m\u00e1y c\u1ea7m pahir t\u1eaft d\u1ea7n nhanh<\/p>\n

C\u00e2u 34:<\/strong>(ID : 84364)<\/strong>Ng\u01b0\u1eddi ta truy\u1ec1n t\u1ea3i \u0111i\u1ec7n xoay chi\u1ec1u m\u1ed9t pha t\u1eeb m\u1ed9t tr\u1ea1m ph\u00e1t \u0111i\u1ec7n c\u00e1ch n\u01a1i ti\u00eau th\u1ee5 10km. D\u00e2y d\u1eabn l\u00e0m b\u1eb1ng kim lo\u1ea1i c\u00f3 \u0111i\u1ec7n tr\u1edf su\u1ea5t 2,5.10<\/p>\n

-8<\/dd>\n

  \u2126m, ti\u1ebft di\u1ec7n 0,4cm<\/p>\n

2<\/dd>\n

, h\u1ec7 s\u1ed1 c\u00f4ng su\u1ea5t c\u1ee7a m\u1ea1ch \u0111i\u1ec7n l\u00e0 0,9. \u0110i\u1ec7n \u00e1p v\u00e0 c\u00f4ng su\u1ea5t truy\u1ec1n \u0111i \u1edf tr\u1ea1m ph\u00e1t \u0111i\u1ec7n l\u00e0 10kV v\u00e0 500kW. Hi\u1ec7u su\u1ea5t truy\u1ec1n t\u1ea3i \u0111i\u1ec7n l\u00e0:<\/p>\n<\/div>\n

A.<\/strong>96,34%<\/span><\/p>\n

\n

B.<\/strong>92,28%<\/p>\n

C.<\/strong>93,75%<\/p>\n

D.<\/strong>96,88%<\/p>\n<\/div>\n

C\u00e2u 35: (ID : 84365)<\/strong>M\u1ed9t m\u00e1y ph\u00e1t thanh v\u00f4 tuy\u1ebfn truy\u1ec1n \u0111\u01a1n gi\u1ea3n nh\u1ea5t c\u1ea7n c\u00e1c b\u1ed9 ph\u1eadn (1) anten ph\u00e1t ;(2) m\u1ea1ch bi\u1ebfn \u0111i\u1ec7u; (3) m\u1ea1ch khu\u1ebfch \u0111\u1ea1i ; (4 m\u1ea1ch ph\u00e1t s\u00f3ng \u0111i\u1ec7n t\u1eed cao t\u1ea7n; (5) micr\u00f4. Th\u1ee9 t\u1ef1  s\u1eafp x\u1ebfp c\u00e1c b\u1ed9 ph\u1eadn tr\u00ean trong m\u00e1y ph\u00e1t thanh l\u00e0:<\/span><\/p>\n

A.(<\/strong>5) \u2013 (4) \u2013(3) \u2013 (2) \u2013 (1)<\/span><\/p>\n

B.<\/strong>.(5) \u2013 (4) \u2013(2) \u2013 (3) \u2013 (1)<\/p>\n

C.<\/strong>.(5) \u2013 (2) \u2013(4) \u2013 (3) \u2013 (1)<\/p>\n

D. .(<\/strong>5) \u2013 (2) \u2013(3) \u2013 (4) \u2013 (1)<\/p>\n

\n

C\u00e2u 36: (ID : 84366)<\/strong> \u0110o\u1ea1n m\u1ea1ch xoay chi\u1ec1u g\u1ed3m cu\u1ed9n d\u00e2y c\u00f3 \u0111i\u1ec7n tr\u1edf thu\u1ea7n R , \u0111\u1ed9 t\u1ef1 c\u1ea3m L m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi t\u1ee5 \u0111i\u1ec7n c\u00f3 dung \u0111i\u1ec7n dung C. \u0110i\u1ec7n \u00e1p \u0111\u1eb7t v\u00e0o hai \u0111\u1ea7u m\u1ea1ch c\u00f3 t\u1ea7n s\u1ed1 f v\u00e0 gi\u00e1 tr\u1ecb hi\u1ec7u d\u1ee5ng 37,5 V. Khi \u0111\u1ed1 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 hi\u1ec7u d\u1ee5ng I = 0,1A ; \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng gi\u1eefa hai \u0111\u1ea7u cu\u1ed9n d\u00e2y l\u00e0 50V, gi\u1eefa hai b\u1ea3n c\u1ef1c t\u1ee5 \u0111i\u1ec7n l\u00e0 17,5V. N\u1ebfu cho t\u1ea7n s\u1ed1 f thay \u0111\u1ed5i \u0111\u1ebfn gi\u00e1 tr\u1ecb f\u2019 = 330 Hz th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n hi\u1ec7u d\u1ee5ng \u0111\u1ea1t gi\u00e1 tr\u1ecb  c\u1ef1c \u0111\u1ea1i .T\u00e2n s\u1ed1  f c\u00f3 gi\u00e1 tr\u1ecb<\/p>\n<\/div>\n

A .<\/strong> \u2248 100 \u03c0  Hz     <\/span><\/p>\n

\n

B .<\/strong> \u2248 500 Hz <\/p>\n

C .<\/strong> \u2248 500 \u03c0    <\/p>\n

A .<\/strong> \u2248 1000Hz      <\/p>\n<\/div>\n

C\u00e2u 37:<\/strong>(ID : 84367)<\/strong>m\u1ed9t ngu\u1ed3n \u00e2m \u0111\u1eb7t t\u1ea1o O ph\u00e1t s\u00f3ng \u0111\u1eb3ng h\u01b0\u1edbng trong kh\u00f4ng gian . A v\u00e0 B l\u00e0 2 \u0111i\u1ec3m nawmfd tr\u00e2n c\u00f9ng 1 tia xu\u1ea5t ph\u00e1t t\u1eeb O.I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a  AB . G\u1ecdi L<\/span><\/p>\n

A<\/dd>\n

, L<\/span><\/p>\n

I<\/dd>\n

, L<\/span><\/p>\n

B<\/dd>\n

L\u1ea7n l\u01b0\u1ee3t l\u00e0 m\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m t\u1ea1i  A, I, B ta c\u00f3 L<\/span><\/p>\n

A<\/dd>\n

-L<\/span><\/p>\n

I<\/dd>\n

= 20 (bB)<\/span><\/p>\n

A .<\/strong> L<\/span><\/p>\n

I<\/dd>\n

-L<\/span><\/p>\n

B<\/dd>\n

= 2,56 B         <\/span><\/p>\n

\n

B .<\/strong> L<\/p>\n

B<\/dd>\n

-L<\/p>\n

I<\/dd>\n

= 0,56 (B )    <\/p>\n

C .<\/strong> L<\/p>\n

A<\/dd>\n

-L<\/p>\n

B<\/dd>\n

= 2,56 (B ) <\/p>\n

D .<\/strong> L<\/p>\n

B<\/dd>\n

-L<\/p>\n

A<\/dd>\n

= 0,56 (B )      <\/p>\n<\/div>\n

C\u00e2u 38 :<\/strong>(ID : 84368)<\/strong>\u0111o\u1ea1n m\u1ea1ch AB n\u1ed1i ti\u1ebfp theo th\u1ee9 t\u1ef1 g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n , cu\u1ed9n d\u00e2y thu\u1ea7n c\u1ea3m v\u00e0 t\u1ee5 \u0111i\u1ec7n c\u00f3 th\u1ec3 thay \u0111\u1ed5i  \u0111i\u1ec7n dung . \u0110i\u1ec3m M n\u1eb1m gi\u1eefa  cu\u1ed9n d\u00e2y va t\u1ee5 . \u0110i\u1ec7n \u00e1p hai \u0111\u1ea7u m\u1ea1ch  l\u00e0  u = 200cos(2 \u03c0ft) (V) . Ban \u0111\u1ea7u \u0111i\u1ec7n \u00e1p gi\u1eefa AM l\u1ec7ch AM l\u1ec7ch pha \u03c0\/2 so v\u1edbi \u0111i\u1ec7n \u00e1p hai \u0111\u1ea7u m\u1ea1ch . T\u0103ng t\u1ea7n s\u1ed1 c\u1ee7a d\u00f2ng \u0111i\u1ec7n l\u00ean 2 l\u1ea7n th\u00ec \u0111i\u1ec7n \u00e1p gi\u1eefa MB<\/span><\/p>\n

A .<\/strong> t\u0103ng 2 l\u1ea7n    <\/span><\/p>\n

\n

B .<\/strong> kh\u00f4ng \u0111\u1ed5i    <\/p>\n

C .<\/strong> t\u0103ng 4 l\u1ea7n    <\/p>\n

D .<\/strong>  gi\u1ea3m<\/p>\n<\/div>\n

C\u00e2u 39 :<\/strong>(ID : 84369)<\/strong>v\u1eadt nh\u1ecf c\u1ee7a m\u1ed9t con l\u1eafc  \u0111\u01a1n c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng l\u01b0\u1ee3ng 20g . Con l\u1eafc dao \u0111\u1ed9ng v\u1edbi chu k\u1ef3 b\u1eb1ng 2s t\u1ea1i n\u01a1i c\u00f3 gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng b\u1eb1ng 9,8 m\/s<\/span><\/p>\n

2<\/dd>\n

.Truy\u1ec1n cho v\u1eadt nh\u1ecf c\u1ee7a  con l\u1eafc  \u0111i\u1ec7n q = 2.10<\/span><\/p>\n

-7<\/dd>\n

C r\u1ed3i  \u0111\u1eb7t n\u00f3 v\u00e0o  \u0111i\u1ec7n  tr\u01b0\u1eddng  \u0111\u1ec1u c\u00f3 tr\u01b0\u1eddng  c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 2,77.10<\/span><\/p>\n

5<\/dd>\n

 V\/m. \u0110\u01b0\u1eddng s\u1ee9c \u0111i\u1ec7n tr\u01b0\u1eddng c\u00f3 ph\u01b0\u01a1ng  ngang , song  song v\u1edbi m\u1eb7t ph\u1eb3ng dao \u0111\u1ed9ng  c\u1ee7a con l\u1eafc . Chu k\u1ef3 c\u1ee7a con l\u1eafc trong \u0111i\u1ec7n tr\u01b0\u1eddng l\u00e0 :<\/span><\/p>\n

A .<\/strong> 1,96s        <\/span><\/p>\n

\n

B .<\/strong> 2,04s    <\/p>\n

C.<\/strong> 14 cm      <\/p>\n

D .<\/strong> 15,46 cm<\/p>\n<\/div>\n

C\u00e2u 40 :<\/strong>(ID : 84370)<\/strong>Trong c\u00e1c nh\u1ea1c c\u1ee5 , h\u1ed9p \u0111\u00e0n c\u00f3 t\u00e1c d\u1ee5ng<\/span><\/p>\n

\n

A .<\/strong> v\u1eeba khuy\u1ebfch  \u0111\u1ea1i c\u01b0\u1eddng \u0111\u1ed9 \u00e2m ,v\u1eeba t\u1ea1o ra \u00e2m s\u1eafc ri\u00eang c\u1ee7a \u00e2m do \u0111\u00e0n ph\u00e1t ra<\/p>\n

B .<\/strong> gi\u1eef cho \u00e2m ph\u00e1t ra c\u00f3 t\u1ea7n s\u1ed1 \u1ed5n \u0111\u1ecbnh nh\u1edd hi\u1ec7n t\u01b0\u1ee3ng c\u1ed9ng h\u01b0\u1edfng<\/p>\n

C .<\/strong> L\u00e0m tri\u1ec7t ti\u00eau nh\u1eefng \u00e2m c\u00f3 t\u00e0n s\u1ed1 kh\u00f4ng mong mu\u1ed1n<\/p>\n

B .<\/strong> tr\u00e1nh \u0111\u01b0\u1ee3c t\u1ea1p \u00e2m v\u00e0 ti\u1ebfng \u1ed3n , l\u00e0m cho ti\u1ebfng \u0111\u00e0n trong tr\u1ebbo<\/p>\n

C\u00e2u 41 : (ID : 84371)<\/strong>trongTN  giao thoa \u00e1nh sang Young : kho\u1ea3ng c\u00e1ch gi\u1eefa hai khe l\u00e0  0,2 mm , kho\u1ea3ng c\u00e1ch t\u1eeb 2 khe \u0111\u1ebfn m\u00e0n l\u00e0 40 cm , b\u01b0\u1edbc s\u00f3ng \u00e1nh s\u00e1ng b\u1eb1ng 0,5 \u03bcm . Trong kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m A, B \u1edf c\u00f9ng ph\u00eda \u0111\u1ed1i t\u01b0\u1ee3ng O v\u00e0 t\u1ecda \u0111\u1ed9 l\u1ea7n l\u01b0\u1ee3t  l\u00e0 X<\/p>\n

M<\/dd>\n

= 2mm , X<\/p>\n

N<\/dd>\n

= 6,5 mm c\u00f3 :<\/p>\n<\/div>\n

A .<\/strong> 5 v\u00e2n s\u00e1ng v\u00e0 4 v\u00e2n t\u1ed1i<\/span><\/p>\n

\n

B .<\/strong> 4 v\u00e2n s\u00e1ng v\u00e0 5 v\u00e2n t\u1ed1i<\/p>\n

C.<\/strong> 5 v\u00e2n s\u00e1ng v\u00e0 5 v\u00e2n t\u1ed1i<\/p>\n

D .<\/strong> 4 v\u00e2n s\u00e1ng v\u00e0 4 v\u00e2n t\u1ed1i<\/p>\n<\/div>\n

C\u00e2u 42 : (ID : 84372)<\/strong>k\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y l\u00e0 sai khi n\u00f3i v\u1ec1 m\u00e1y quang ph\u1ed5 ? M\u00e1y quang ph\u1ed5<\/span><\/p>\n

\n

A .<\/strong> l\u00e0 d\u1ee5ng c\u1ee5 d\u00f9ng \u0111\u1ec1 ph\u00e2n t\u00edch chum \u00e1nh sang ph\u1ee9c t\u1ea1p  th\u00e0nh nh\u1eefng th\u00e0nh ph\u1ea7n \u0111\u01a1n s\u1eafc<\/p>\n

B .<\/strong> c\u00f3 ba b\u1ed9 ph\u1eadn ch\u00ednh : \u1ed1ng chu\u1ea9n tr\u1ef1c , h\u1ec7 t\u00e1n s\u1eafc v\u00e0 bu\u1ed3ng \u1ea3nh<\/p>\n

C.<\/strong>c\u00f3 khe F cu\u1ea3 \u1ed1ng chu\u1ea9n tr\u1ef1c tr\u00f9ng v\u1edbi ti\u00eau \u0111i\u1ec3m ch\u00ednh c\u1ee7a th\u1ea5u k\u00ednh ph\u00e2n k\u00ec<\/p>\n

D.  <\/strong>ho\u1ea1t  \u0111\u1ed9ng d\u1ef1a tr\u00ean hi\u1ec7n t\u01b0\u1ee3ng t\u00e1n s\u1eafc \u00e1nh s\u00e1ng<\/p>\n

C\u00e2u 43 : (ID : 84373)<\/strong>tr\u00ean m\u1eb7t n\u01b0\u1edbc c\u00f3 hai ngu\u1ed3n k\u1ebft h\u1ee3p O<\/p>\n

1<\/dd>\n

,O<\/p>\n

2<\/dd>\n

, c\u00e1ch nhau 8 cm dao \u0111\u1ed9ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u1ee9ng l\u00e0 : u<\/p>\n

1<\/dd>\n

= 2cos(40 \u03c0t \u2013 \u03c0\/6) cm , u<\/p>\n

2<\/dd>\n

= 2cos(40 \u03c0t + 5\u03c0\/6) cm. V\u1eadn t\u1ed1c truy\u1ec1n s\u00f3ng l\u00e0 v = 40 cm\/s . S\u1ed1 \u0111i\u1ec3m dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 2 cm tr\u00ean \u0111o\u1ea1n O<\/p>\n

1<\/dd>\n

O<\/p>\n

2<\/dd>\n

l\u00e0<\/p>\n<\/div>\n

A .<\/strong> 16    <\/span><\/p>\n

\n

B .<\/strong> 8    <\/p>\n

C.<\/strong>9    <\/p>\n

D. 18<\/strong><\/p>\n<\/div>\n

C\u00e2u 44 : (ID : 84374)<\/strong>M\u1ed9t s\u00f3ng lan truy\u1ec1n tr\u00ean m\u1ed9t s\u1ee3i d\u00e2y v\u1edbi bi\u00ean \u0111\u1ed9 kh\u00f4ng \u0111\u1ed5i . T\u1ea1i th\u1eddi \u0111i\u1ec3m t =0 , ph\u1ea7n  t\u1eed m\u00f4i tr\u01b0\u1eddng t\u1ea1i ngu\u1ed3n ph\u00e1t \u0111i qua v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng theo chi\u1ec1u d\u01b0\u01a1ng . \u0110i\u1ec3m M c\u00e1ch ngu\u1ed3n m\u1ed9t kho\u1ea3ng 1\/6 b\u01b0\u1edbc  s\u00f3ng c\u00f3 li \u0111\u1ed9 3 cm t\u1ea1i th\u1eddi \u0111i\u1ec3m \u0111i\u1ec3m t = T\/4 . Bi\u00ean \u0111\u1ed9 s\u00f3ng l\u00e0 :<\/span><\/p>\n

\n

A .<\/strong> 4cm    B .<\/strong> 6 cm     C.<\/strong>5 cm   D.<\/strong>3cm<\/p>\n

C\u00e2u 45 : (ID : 84375)<\/strong>M\u1ed9t m\u1ea1ch dao \u0111\u1ed9ng LC l\u00ed t\u01b0\u1edfng c\u00f3 L = 2.10<\/p>\n

-6<\/dd>\n

H, C= 8.10<\/p>\n

-8<\/dd>\n

F v\u00e0 c\u01b0\u1eddng \u0111\u1ed9 c\u1ef1c \u0111\u1ea1i c\u1ee7a d\u00f2ng  \u0111i\u1ec7n ch\u1ea1y trong m\u1ea1ch I<\/p>\n

0<\/dd>\n

= 0,5A. L\u1ea5y g\u1ed1c th\u1eddi gian t = 0 l\u00e0 l\u00fac n\u0103ng l\u01b0\u1ee3ng \u0111i\u1ec7n tr\u01b0\u1eddng b\u1eb1ng n\u0103ng l\u01b0\u1ee3ng t\u1eeb tr\u01b0\u1eddng , \u0111i\u1ec7n t\u00edch c\u1ee7a m\u1ed9t b\u1ea3n t\u1ee5 \u0111i\u1ec7n v\u00e0 c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n  ch\u1ea1y trong m\u1ea1ch \u0111\u1ec1u  c\u00f3 gi\u00e1 tr\u1ecb d\u01b0\u01a1ng . \u0111i\u1ec7n t\u00edch tr\u00ean t\u1ee5 bi\u1ebfn thi\u00ean theo quy lu\u1eadt:<\/p>\n<\/div>\n

A .<\/strong> q =10<\/span><\/p>\n

-7<\/dd>\n

cos(2,5.10<\/span><\/p>\n

6<\/dd>\n

t \u2013 \u03c0\/4) (C)                                                     <\/span><\/p>\n

B .<\/strong> q =10<\/p>\n

-7<\/dd>\n

cos(2,5.10<\/p>\n

6<\/dd>\n

t +  \u03c0\/4) (C)<\/p>\n

C.<\/strong>q =10<\/p>\n

-7<\/dd>\n

cos(2,5.10<\/p>\n

6<\/dd>\n

t + \u03c0\/3) (C)<\/p>\n

D.<\/strong>q =10<\/p>\n

-7<\/dd>\n

cos(2,5.10<\/p>\n

6<\/dd>\n

t \u2013 \u03c0\/3) (C)<\/p>\n

\n

C\u00e2u 46 :<\/strong> (ID : 84376)<\/strong>tr\u00ean m\u1ed9t s\u1ee3i d\u00e2y \u0111ang c\u00f3 s\u00f3ng d\u1eebng v\u1edbi b\u01b0\u1edbc s\u00f3ng \u03bb . A l\u00e0 m\u1ed9t \u0111i\u1ec3m n\u00fat , B l\u00e0 m\u1ed9t \u0111i\u1ec3m b\u1ee5ng v\u00e0 C l\u00e0 \u0111i\u1ec3m n\u1eb1m trong kho\u1ea3ng A,B m\u00e0 trong 1 chu k\u00ec , th\u1eddi gian li \u0111\u1ed9 c\u1ee7a B c\u00f3 \u0111\u1ed9 l\u1edbn h\u01a1n bi\u00ean \u0111\u1ed9 c\u1ee7a C l\u00e0 T\/3 .Kho\u1ea3ng c\u00e1ch AC b\u1eb1ng<\/p>\n<\/div>\n

A<\/strong>.  \u03bb\/12<\/span><\/p>\n

\n

B .<\/strong>  \u03bb\/6<\/p>\n

C.<\/strong>\u03bb\/4<\/p>\n

D.<\/strong>\u03bb\/3<\/p>\n<\/div>\n

C\u00e2u 47 : (ID : 84377)<\/strong>con l\u1eafc n\u1eb1m ngang g\u1ed3m v\u1eadt nh\u1ecf c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng 100g v\u00e0 lo xo c\u00f3 \u0111\u1ed9 c\u1ee9ng 10N\/m \u0111\u1eb7t tr\u00ean m\u1eb7t ph\u1eb3ng c\u00f3 h\u1ec7 s\u1ed1 ma s\u00e1t  b\u1eb1ng 0,2 . L\u1ea5y g = 10m\/s<\/span><\/p>\n

2<\/dd>\n

.\u0110\u01b0a v\u1eadt t\u1edbi v\u1ecb tr\u00ed l\u00f2 xo n\u00e9n 10 cm r\u1ed3i th\u1ea3 nh\u1eb9 .Ngay sau khi th\u1ea3 v\u1eadt , n\u00f3 chuy\u1ec3n \u0111\u1ed9ng theo chi\u1ec1u d\u01b0\u01a1ng .T\u1ed1c \u0111\u1ed9 c\u1ef1c \u0111\u1ea1i c\u1ee7a v\u1eadt nh\u1ecf trong qu\u00e1 tr\u00ecnh n\u00f3 chuy\u1ec3n \u0111\u1ed9ng theo chi\u1ec1u \u00e2m \u0111\u1ea7u ti\u00ean l\u00e0 :<\/span><\/p>\n

A .<\/strong> 0,35 m\/s<\/span><\/p>\n

\n

B .<\/strong> 0,4m\/s<\/p>\n

C.<\/strong>0,7m\/s<\/p>\n

D.<\/strong>0,8m\/s<\/p>\n<\/div>\n

C\u00e2u 48 : (ID : 84378)<\/strong>\u1eed d\u1ee5ng \u00e1nh s\u00e1ng tr\u1eafng co b\u01b0\u1edbc song t\u1eeb 0,38 \u03bcm \u0111\u1ebfn 0,76 \u03bcm trong th\u00ed nghi\u1ec7m Young. Bi\u1ebft kho\u1ea3ng c\u00e1ch gi\u1eefa hai khe l\u00e0 0,1 mm, kho\u1ea3ng c\u00e1ch t\u1eeb hai khe t\u1edbi m\u00e0n l\u00e0 30cm . C\u00e1c b\u1ee9c x\u1ea1 cho c\u1ef1c \u0111\u1ea1i giao thoa t\u1ea1i v\u1ecb tr\u00ed c\u1ef1c ti\u1ec3u giao thoa th\u1ee9 ba c\u1ee7a b\u1ee9c x\u1ea1 c\u00f3 b\u01b0\u1edbc s\u00f3ng  0,5 \u03bcm l\u00e0<\/span><\/p>\n

A .<\/strong> 0,625 \u03bcm<\/span><\/p>\n

\n

B .<\/strong> 0,42 \u03bcm v\u00e0 0,5 \u03bcm<\/p>\n

C. <\/strong>0,58 \u03bcm v\u00e0 0,44 \u03bcm<\/p>\n

D.<\/strong> 0,58 \u03bcm<\/p>\n<\/div>\n

C\u00e2u 49 : (ID : 84379)<\/strong>khi con l\u1eafc l\u00f2 xo th\u1eb3ng \u0111\u1ee9ng \u1edf v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng th\u00ec l\u00f2 xo d\u00e3n 5 cm .L\u1ea5y g =10m\/s<\/span><\/p>\n

2<\/dd>\n

.Bi\u1ebft r\u1eb1ng trong m\u1ed9t chu k\u1ef3 , th\u1eddi gian l\u00f2 xo n\u00e9n b\u1eb1ng th\u1eddi gian l\u00f2 xo gi\u00e3n .T\u1ed1c \u0111\u1ed9 c\u1ee7a v\u1eadt khi \u0111i qua v\u1ecb tr\u00ed l\u00f2 xo kh\u00f4ng bi\u1ebfn d\u1ea1ng l\u00e0 :<\/span><\/p>\n

A . <\/strong> \u221a2\/2m\/s<\/span><\/p>\n

\n

B .<\/strong> \u221a3\/\u221a2 m\/s<\/p>\n

C.<\/strong>\u221a6\/2 m\/s<\/p>\n

D.<\/strong>\u221a3\/2 m\/s<\/p>\n<\/div>\n

C\u00e2u 50 : (ID : 84380)<\/strong>cho l\u0103ng k\u00ednh c\u00f3 g\u00f3c tri\u1ebft quang A = 45<\/span><\/p>\n

0<\/dd>\n

\u0111\u1eb7t trong kh\u00f4ng kh\u00ed . N\u1ebfu chi\u1ebfu chum tia sang song song h\u1eb9p \u0111\u01a1n s\u1eafc m\u00e0u l\u1ee5c  t\u1edbi vu\u00f4ng g\u00f3c m\u1eb7t b\u00ean AB c\u1ee7a l\u0103ng k\u00ednh th\u00ec tia l\u00f3 ra kh\u1ecfi l\u0103ng k\u00ednh \u0111i s\u00e1t m\u1eb7t b\u00ean AC. N\u1ebfu chi\u1ebfu 1 ch\u00f9m tia song song h\u1eb9p g\u1ed3m 4 \u00e1nh sang \u0111\u01a1n s\u1eafc :lam , ch\u00e0m , cam , \u0111\u1ecf t\u1edbi vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t b\u00ean AB th\u00ec kh\u00f4ng l\u00f3 ra m\u1eb7t b\u00ean AC  l\u00e0 :<\/span><\/p>\n

A .<\/strong> 2 tia:m\u00e0u lam v\u00e0  m\u00e0u \u0111\u1ecf<\/span><\/p>\n

B .<\/strong> 2 tia : m\u00e0u ch\u00e0m v\u00e0 m\u00e0u lam<\/p>\n

C .<\/strong> 2 tia : m\u00e0u \u0111\u1ecf v\u00e0 m\u00e0u cam<\/p>\n

D.<\/strong>2 tia :m\u00e0u cam v\u00e0 m\u00e0u ch\u00e0m <\/p>\n

\u0110\u00e1p \u00e1n \u0111\u1ec1 thi th\u1eed  THPT Qu\u1ed1c gia m\u00f4n V\u1eadt L\u00fd n\u0103m 2016- \u0110\u1ec1 s\u1ed1 7<\/strong><\/span><\/p>\n

<\/strong><\/span><\/strong><\/span><\/p>\n<\/p>\n

<\/strong><\/span><\/p>\n

T\u1ea5t c\u1ea3 n\u1ed9i dung b\u00e0i vi\u1ebft. C\u00e1c em h\u00e3y xem th\u00eam v\u00e0 t\u1ea3i file chi ti\u1ebft t\u1ea1i \u0111\u00e2y:<\/strong>Download<\/p>\n","protected":false},"excerpt":{"rendered":"

Tham kh\u1ea3o \u0111\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n V\u1eadt L\u00fd \u0111\u1ec1 s\u1ed1 7 c\u00f3 \u0111\u00e1p \u00e1n v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft gi\u00fap c\u00e1c em th\u1eed s\u1ee9c v\u1edbi k\u1ef3 thi s\u1eafp t\u1edbi. Xem th\u00eam: \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n V\u1eadt L\u00fd n\u0103m 2016 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[25],"tags":[],"yoast_head":"\n