C\u1eadp nh\u1eadt \u0111\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n v\u1eadt l\u00fd n\u0103m 2016 gi\u00fap c\u00e1c em l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 thi THPT Qu\u1ed1c gia s\u1eafp t\u1edbi.<\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd n\u0103m 2016 – \u0110\u1ec1 s\u1ed1 1<\/strong> <\/p>\n

\n

C\u00e2u 1<\/strong>(ID: 82547)<\/strong>:<\/strong>S\u00f3ng d\u1eebng xu\u1ea5t hi\u1ec7n tr\u00ean m\u1ed9t s\u1ee3i d\u00e2y v\u1edbi t\u1ea7n s\u1ed1 5 Hz. G\u1ecdi th\u1ee9 t\u1ef1 c\u00e1c \u0111i\u1ec3m thu\u1ed9c d\u00e2y l\u1ea7n l\u01b0\u1ee3t l\u00e0 O, M, N, P sao cho O l\u00e0 \u0111i\u1ec3m n\u00fat, P l\u00e0 \u0111i\u1ec3m b\u1ee5ng s\u00f3ng g\u1ea7n O nh\u1ea5t (M, N thu\u1ed9c \u0111o\u1ea1n OP). Kho\u1ea3ng th\u1eddi gian gi\u1eefa 2 l\u1ea7n li\u00ean ti\u1ebfp \u0111\u1ec3 \u0111\u1ed9 l\u1edbn li \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m P b\u1eb1ng bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng c\u1ee7a \u0111i\u1ec3m M v\u00e0 N l\u1ea7n l\u01b0\u1ee3t l\u00e0 1\/20 (s) v\u00e0 1\/15 (s). Bi\u1ebft kho\u1ea3ng c\u00e1ch gi\u1eefa 2 \u0111i\u1ec3m M v\u00e0 N l\u00e0 0,2 cm. S\u00f3ng truy\u1ec1n tr\u00ean d\u00e2y c\u00f3 b\u01b0\u1edbc s\u00f3ng g\u1ea7n gi\u00e1 tr\u1ecb n\u00e0o nh\u1ea5t?<\/p>\n<\/div>\n

\n

A<\/strong>. 1 cm     <\/p>\n

B.<\/strong>4 cm     <\/p>\n

C.<\/strong>5 cm     <\/p>\n

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

\n

C\u00e2u 2<\/strong>(ID: 82551)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p u = U<\/p>\n

0<\/dd>\n

cos(\u03c9t + \u03c0\/4) v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch ch\u1ec9 c\u00f3 t\u1ee5 \u0111i\u1ec7n th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch l\u00e0 I = I<\/p>\n

0<\/dd>\n

cos(\u03c9t + \u03c6). Gi\u00e1 tr\u1ecb c\u1ee7a \u03c6 b\u1eb1ng :<\/p>\n<\/div>\n

\n

A.<\/strong>3\u03c0\/4        <\/p>\n

B.<\/strong>-3\u03c0\/4    <\/p>\n

C.<\/strong>\u2013\u03c0\/2       <\/p>\n

D.<\/strong>\u03c0\/2<\/p>\n<\/div>\n

\n

C\u00e2u 3<\/strong>(ID: 82553)<\/strong>:<\/strong>M\u1ed9t s\u00f3ng d\u1eebng tr\u00ean s\u1ee3i d\u00e2y th\u1eb3ng d\u00e0i n\u1eb1m d\u1ecdc tr\u1ee5c Ox c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh u = 2sin(\u03c0x\/4)cos20\u03c0t (x \u0111o b\u1eb1ng cm, t \u0111o b\u1eb1ng s). C\u00e1c n\u00fat s\u00f3ng c\u00f3 to\u1ea1 \u0111\u1ed9 x<\/p>\n

1<\/dd>\n

, c\u00e1c b\u1ee5ng s\u00f3ng c\u00f3 to\u1ea1 \u0111\u1ed9 x<\/p>\n

2<\/dd>\n

(x<\/p>\n

1<\/dd>\n

, x<\/p>\n

2<\/dd>\n

\u0111o b\u1eb1ng cm, k nguy\u00ean) c\u00f3 gi\u00e1 tr\u1ecb t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>x<\/p>\n

1<\/dd>\n

= 4k; x<\/p>\n

2<\/dd>\n

= 2 + 4k       <\/p>\n

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

1<\/dd>\n

= 2k ; x<\/p>\n

2<\/dd>\n

= 2k + 1<\/p>\n

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

1<\/dd>\n

= 2k + 1; x<\/p>\n

2<\/dd>\n

= 4k        <\/p>\n

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

1<\/dd>\n

= 8k ; x<\/p>\n

2<\/dd>\n

= 2k + 1<\/p>\n<\/div>\n

\n

C\u00e2u 4<\/strong>(ID: 82554)<\/strong>:<\/strong>T\u1ea7n s\u1ed1 dao \u0111\u1ed9ng ri\u00eang c\u1ee7a m\u1ed9t con l\u1eafc l\u00f2 xo l\u00e0 f<\/p>\n

0<\/dd>\n

. Ngo\u1ea1i l\u1ef1c t\u00e1c d\u1ee5ng v\u00e0o con l\u1eafc c\u00f3 d\u1ea1ng: F = F<\/p>\n

0<\/dd>\n

cos2\u03c0ft (F<\/p>\n

0<\/dd>\n

kh\u00f4ng \u0111\u1ed5i, f thay \u0111\u1ed5i \u0111\u01b0\u1ee3c). G\u1ecdi A<\/p>\n

0<\/dd>\n

, A<\/p>\n

1<\/dd>\n

, A<\/p>\n

2<\/dd>\n

l\u00e0 bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng c\u1ee7a con l\u1eafc n\u00e0y t\u01b0\u01a1ng \u1ee9ng v\u1edbi c\u00e1c t\u1ea7n s\u1ed1 khi f = f<\/p>\n

0<\/dd>\n

; f = f<\/p>\n

1<\/dd>\n

, f = f<\/p>\n

2<\/dd>\n

. Bi\u1ebft f<\/p>\n

2<\/dd>\n

= 2f<\/p>\n

1<\/dd>\n

< f<\/p>\n

0<\/dd>\n

. Li\u00ean h\u1ec7 \u0111\u00fang l\u00e0:<\/p>\n<\/div>\n

\n

A.<\/strong>A2 > A1      <\/p>\n

B.<\/strong>A2 < A1     <\/p>\n

C.<\/strong>A2 = A1     <\/p>\n

D.<\/strong>A2 = A0<\/p>\n<\/div>\n

\n

C\u00e2u 5<\/strong>(ID: 82555)<\/strong>:<\/strong>M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 v\u1edbi t\u1ea7n s\u1ed1 f v\u00e0 bi\u00ean \u0111\u1ed9 A. Th\u1eddi gian v\u1eadt \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 2A l\u00e0:<\/p>\n<\/div>\n

\n

A.<\/strong>1\/3f        <\/p>\n

B.<\/strong>1\/12f          <\/p>\n

C.<\/strong>1\/4f        <\/p>\n

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

\n

C\u00e2u 6<\/strong>(ID: 82566)<\/strong>:<\/strong>M\u1ed9t m\u00e1y bi\u1ebfn \u00e1p l\u00ed t\u01b0\u1edfng c\u00f3 s\u1ed1 v\u00f2ng d\u00e2y cu\u1ed9n s\u01a1 c\u1ea5p g\u1ea5p 10 l\u1ea7n s\u1ed1 v\u00f2ng d\u00e2y cu\u1ed9n th\u1ee9 c\u1ea5p. M\u00e1y bi\u1ebfn \u00e1p n\u00e0y<\/p>\n

A.<\/strong>l\u00e0 m\u00e1y t\u0103ng \u00e1p<\/p>\n

B.<\/strong>l\u00e0m gi\u1ea3m t\u1ea7n s\u1ed1 d\u00f2ng \u0111i\u1ec7n \u1edf cu\u1ed9n s\u01a1 c\u1ea5p 10 l\u1ea7n<\/p>\n

C.<\/strong>l\u00e0 m\u00e1y h\u1ea1 \u00e1p<\/p>\n

D.<\/strong>l\u00e0m t\u0103ng t\u1ea7n s\u1ed1 d\u00f2ng \u0111i\u1ec7n \u1edf cu\u1ed9n s\u01a1 c\u1ea5p 10 l\u1ea7n<\/p>\n

C\u00e2u 7<\/strong>(ID: 82567)<\/strong>:<\/strong>M\u1ed9t m\u00e1y bi\u1ebfn \u00e1p l\u00ed t\u01b0\u1edfng, cu\u1ed9n s\u01a1 c\u1ea5p c\u00f3 N<\/p>\n

1<\/dd>\n

= 1100 v\u00f2ng \u0111\u01b0\u1ee3c n\u1ed1i v\u00e0o \u0111i\u1ec7n \u00e1p xoay chi\u1ec1u c\u00f3 gi\u00e1 tr\u1ecb hi\u1ec7u d\u1ee5ng 220V. th\u1ee9 c\u1ea5p g\u1ed3m hai cu\u1ed9n : N<\/p>\n

2<\/dd>\n

= 55 v\u00f2ng, N<\/p>\n

3<\/dd>\n

= 110 v\u00f2ng. Gi\u1eefa hai \u0111\u1ea7u N<\/p>\n

2<\/dd>\n

\u0111\u1ea5u v\u1edbi \u0111i\u1ec7n tr\u1edf thu\u1ea7n R1 = 11 \u2126, gi\u1eefa hai \u0111\u1ea7u N<\/p>\n

3<\/dd>\n

\u0111\u1ea5u v\u1edbi \u0111i\u1ec7n tr\u1edf thu\u1ea7n R<\/p>\n

2<\/dd>\n

= 44\u2126. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n hi\u1ec7u d\u1ee5ng ch\u1ea1y trong cu\u1ed9n s\u01a1 c\u1ea5p b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>  0,1A       <\/p>\n

B.<\/strong>0,25A       <\/p>\n

C<\/strong>. 0,05A     <\/p>\n

D.<\/strong>0,15A<\/p>\n<\/div>\n

\n

C\u00e2u 8<\/strong>(ID: 82584)<\/strong>:<\/strong>M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 c\u00f3 chu k\u00ec T. N\u1ebfu ch\u1ecdn g\u1ed1c th\u1eddi gian t = 0 l\u00fac v\u1eadt qua v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng th\u00ec trong n\u1eeda chu k\u00ec \u0111\u1ea7u ti\u00ean, v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt b\u1eb1ng kh\u00f4ng \u1edf th\u1eddi \u0111i\u1ec3m<\/p>\n<\/div>\n

A.<\/strong>  T\/8                <\/p>\n

\n

B.<\/strong>T\/6        <\/p>\n

C.<\/strong>T\/2      <\/p>\n

D.<\/strong>T\/4<\/p>\n<\/div>\n

\n

C\u00e2u 9<\/strong>(ID: 82585)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p u  =120\u221a2cos100\u03c0t (u t\u00ednh b\u1eb1ng V, t t\u00ednh b\u1eb1ng s) v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n 30 \u2126, cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m 2\/5\u03c0 (H) v\u00e0 t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung thay \u0111\u1ed5i \u0111\u01b0\u1ee3c. Khi thay \u0111\u1ed5i \u0111i\u1ec7n dung c\u1ee7a t\u1ee5 \u0111i\u1ec7n th\u00ec \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng gi\u1eefa hai \u0111\u1ea7u t\u1ee5 \u0111i\u1ec7n \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>200V       <\/p>\n

B.<\/strong>120V       <\/p>\n

C.<\/strong>250V     <\/p>\n

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

\n

C\u00e2u 10<\/strong>(ID: 82586)<\/strong>:<\/strong>Khi n\u00f3i v\u1ec1 dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 c\u1ee7a con l\u1eafc l\u00f2 xo n\u1eb1m ngang, ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n

A. <\/strong>V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 tri\u1ec7t ti\u00eau khi qua v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng<\/p>\n

B.  <\/strong>Gia t\u1ed1c c\u1ee7a v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i \u1edf v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng<\/p>\n

C. <\/strong>L\u1ef1c \u0111\u00e0n h\u1ed3i t\u00e1c d\u1ee5ng l\u00ean v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 lu\u00f4n h\u01b0\u1edbng v\u1ec1 v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng.<\/p>\n

D. <\/strong>Gia t\u1ed1c c\u1ee7a v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 tri\u1ec7t ti\u00eau khi \u1edf v\u1ecb tr\u00ed bi\u00ean.<\/p>\n

C\u00e2u 11<\/strong>(ID: 82587)<\/strong>:<\/strong>T\u1ea1i m\u1eb7t ch\u1ea5t l\u1ecfng c\u00f3 4 \u0111i\u1ec3m th\u1eb3ng h\u00e0ng \u0111\u01b0\u1ee3c s\u1eafp x\u1ebfp theo th\u1ee9 t\u1ef1 A,B,C,D v\u1edbi AB = 350 mm; BC = 105 mm, CD = 195 mm. \u0110i\u1ec3m M thu\u1ed9c m\u1eb7t ch\u1ea5t l\u1ecfng c\u00e1ch A v\u00e0 C t\u01b0\u01a1ng \u1ee9ng l\u00e0 MA = 273 mm; MC = 364 mm. Hai ngu\u1ed3n dao \u0111\u1ed9ng theo ph\u01b0\u01a1ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t n\u01b0\u1edbc v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh: u<\/p>\n

1<\/dd>\n

= 3cos100\u03c0t (cm); u<\/p>\n

2<\/dd>\n

= 4cos100\u03c0t (cm). Bi\u1ebft v\u1eadn t\u1ed1c truy\u1ec1n s\u00f3ng tr\u00ean m\u1eb7t ch\u1ea5t l\u1ecfng b\u1eb1ng 12,3 m\/s. Coi bi\u00ean \u0111\u1ed9 s\u00f3ng do c\u00e1c ngu\u1ed3n t\u1edbi M b\u1eb1ng bi\u00ean \u0111\u1ed9 s\u00f3ng c\u1ee7a m\u1ed7i ngu\u1ed3n. Khi hai ngu\u1ed3n s\u00f3ng \u0111\u1eb7t \u1edf A v\u00e0 C th\u00ec c\u00e1c ph\u1ea7n t\u1eed ch\u1ea5t l\u1ecfng t\u1ea1i M dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 A<\/p>\n

1<\/dd>\n

, khi hai ngu\u1ed3n s\u00f3ng \u0111\u1eb7t \u1edf B v\u00e0 D th\u00ec c\u00e1c ph\u1ea7n t\u1eed ch\u1ea5t l\u1ecfng t\u1ea1i M dao \u0111\u1ed9ng v\u1edbi bi\u00ean \u0111\u1ed9 A<\/p>\n

2<\/dd>\n

. Gi\u00e1 tr\u1ecb c\u1ee7a A<\/p>\n

1<\/dd>\n

v\u00e0 A<\/p>\n

2<\/dd>\n

t\u01b0\u01a1ng \u1ee9ng l\u00e0<\/p>\n<\/div>\n

\n

A. <\/strong>2,93 cm v\u00e0 7 cm        <\/p>\n

B.<\/strong>6 cm v\u00e0 2,93 cm<\/p>\n

C.<\/strong>5,1 cm v\u00e0 1,41 cm        <\/p>\n

D.<\/strong>2,93 cm v\u00e0 6,93 cm<\/p>\n<\/div>\n

\n

C\u00e2u 12<\/strong>(ID: 82588)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p u  = U<\/p>\n

0<\/dd>\n

cos100\u03c0t (V) (U<\/p>\n

0<\/dd>\n

kh\u00f4ng \u0111\u1ed5i) v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C, cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L thay \u0111\u1ed5i \u0111\u01b0\u1ee3c. Khi L = L<\/p>\n

1<\/dd>\n

\u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m c\u00f3 gi\u00e1 tr\u1ecb U<\/p>\n

L max<\/dd>\n

v\u00e0 \u0111i\u1ec7n \u00e1p hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch s\u1edbm pha h\u01a1n c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch 0,235\u03b1 ( 0 < \u03b1 < \u03c0\/2). Khi L = L<\/p>\n

2<\/dd>\n

\u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m c\u00f3 gi\u00e1 tr\u1ecb 0,5U<\/p>\n

L max<\/dd>\n

v\u00e0 \u0111i\u1ec7n \u00e1p hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch s\u1edbm pha h\u01a1n so v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n l\u00e0 \u03b1. Gi\u00e1 tr\u1ecb \u03b1 g\u1ea7n gi\u00e1 tr\u1ecb n\u00e0o nh\u1ea5t?<\/p>\n<\/div>\n

\n

A.<\/strong>0,26 rad       <\/p>\n

B.<\/strong>1,36 rad       <\/p>\n

C.<\/strong>0,86 rad     <\/p>\n

D.<\/strong>1,55 rad<\/p>\n<\/div>\n

\n

C\u00e2u 13<\/strong>(ID: 82589)<\/strong>:<\/strong>M\u1ed9t ngu\u1ed3n s\u00e1ng \u0111i\u1ec3m A thu\u1ed9c tr\u1ee5c ch\u00ednh c\u1ee7a m\u1ed9t th\u1ea5u k\u00ednh m\u1ecfng, c\u00e1ch quang t\u00e2m c\u1ee7a th\u1ea5u k\u00ednh 18 cm, qua th\u1ea5u k\u00ednh cho \u1ea3nh A\u2019. Ch\u1ecdn tr\u1ee5c to\u1ea1 \u0111\u1ed9 Ox v\u00e0 O\u2019x\u2019 vu\u00f4ng g\u00f3c v\u1edbi tr\u1ee5c ch\u00ednh c\u1ee7a th\u1ea5u k\u00ednh, c\u00f3 c\u00f9ng chi\u1ec1u d\u01b0\u01a1ng, g\u1ed1c O v\u00e0 O\u2019 thu\u1ed9c tr\u1ee5c ch\u00ednh. Bi\u1ebft Ox \u0111i qua A v\u00e0 O\u2019x\u2019 \u0111i qua A\u2019. Khi A dao \u0111\u1ed9ng tr\u00ean tr\u1ee5c Ox v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x = 4cos(5\u03c0t + \u03c0) (cm) th\u00ec A\u2019 dao \u0111\u1ed9ng tr\u00ean tr\u1ee5c O\u2019x\u2019 v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x\u2019 = 2cos(5\u03c0t + \u03c0) (cm). Ti\u00eau c\u1ef1 c\u1ee7a th\u1ea5u k\u00ednh l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>-9 cm        <\/p>\n

B<\/strong>. 18 cm        <\/p>\n

C.<\/strong>-18 cm       <\/p>\n

D.<\/strong>9 cm<\/p>\n<\/div>\n

\n

C\u00e2u 14<\/strong>(ID: 82590)<\/strong>:<\/strong>Ch\u1ea5t \u0111i\u1ec3m dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 s\u1ebd \u0111\u1ed5i chi\u1ec1u chuy\u1ec3n \u0111\u1ed9ng khi l\u1ef1c k\u00e9o v\u1ec1<\/p>\n<\/div>\n

\n

A.<\/strong>c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c ti\u1ec3u     <\/p>\n

B.<\/strong>c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c \u0111\u1ea1i   <\/p>\n

C.<\/strong>b\u1eb1ng kh\u00f4ng        <\/p>\n

D.<\/strong>\u0111\u1ed5i chi\u1ec1u<\/p>\n<\/div>\n

\n

C\u00e2u 15<\/strong>(ID: 82591)<\/strong>:<\/strong>M\u1ed9t m\u1ea1ch dao \u0111\u1ed9ng LC l\u00ed t\u01b0\u1edfng \u0111ang c\u00f3 dao \u0111\u1ed9ng \u0111i\u1ec7n t\u1eeb v\u1edbi \u0111i\u1ec7n t\u00edch c\u1ef1c \u0111\u1ea1i tr\u00ean hai b\u1ea3n t\u1ee5 l\u00e0 Q0 v\u00e0 d\u00f2ng \u0111i\u1ec7n c\u1ef1c \u0111\u1ea1i trong m\u1ea1ch l\u00e0 I0. Chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a m\u1ea1ch n\u00e0y l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>4\u03c0.Q<\/p>\n

0<\/dd>\n

\/I<\/p>\n

0<\/dd>\n

     <\/p>\n

B.<\/strong>\u03c0.Q<\/p>\n

0<\/dd>\n

\/2I<\/p>\n

0<\/dd>\n

     <\/p>\n

C.<\/strong>2\u03c0.Q<\/p>\n

0<\/dd>\n

\/I<\/p>\n

0<\/dd>\n

      <\/p>\n

D.<\/strong>3\u03c0.Q<\/p>\n

0<\/dd>\n

\/2I<\/p>\n

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

\n

C\u00e2u 16<\/strong>(ID: 82592)<\/strong>:<\/strong>M\u1ed9t ch\u1ea5t \u0111i\u1ec3m dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 kh\u00f4ng ma s\u00e1t tr\u00ean tr\u1ee5c Ox, m\u1ed1c th\u1ebf n\u0103ng \u1edf v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng O. Bi\u1ebft trong qu\u00e1 tr\u00ecnh kh\u1ea3o s\u00e1t ch\u1ea5t \u0111i\u1ec3m kh\u00f4ng \u0111\u1ed5i chi\u1ec1u chuy\u1ec3n \u0111\u1ed9ng. Khi v\u1eeba r\u1eddi kh\u1ecfi v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng m\u1ed9t \u0111o\u1ea1n s th\u00ec \u0111\u1ed9ng n\u0103ng c\u1ee7a ch\u1ea5t \u0111i\u1ec3m l\u00e0 13,95 mJ, \u0111i ti\u1ebfp m\u1ed9t \u0111o\u1ea1n s n\u1eefa th\u00ec \u0111\u1ed9ng n\u0103ng c\u1ee7a ch\u1ea5t \u0111i\u1ec3m ch\u1ec9 c\u00f2n 12,60 mJ. N\u1ebfu ch\u1ea5t \u0111i\u1ec3m \u0111i ti\u1ebfp m\u1ed9t \u0111o\u1ea1n s n\u1eefa th\u00ec \u0111\u1ed9ng n\u0103ng c\u1ee7a ch\u1ea5t \u0111i\u1ec3m khi \u0111\u00f3 b\u1eb1ng<\/p>\n<\/div>\n

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

\n

A.<\/strong>6,68 mJ     <\/p>\n

B.<\/strong>10,35 mJ      <\/p>\n

C.<\/strong>11,25 mJ     <\/p>\n

D<\/strong>. 8,95 mJ<\/p>\n<\/div>\n

\n

C\u00e2u 17<\/strong>(ID: 82593)<\/strong>:<\/strong>Trong gi\u1edd h\u1ecdc th\u1ef1c h\u00e0nh, m\u1ed9t h\u1ecdc sinh c\u1ea7n x\u00e1c \u0111\u1ecbnh sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i \u2206F c\u1ee7a m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng F \u0111o gi\u00e1n ti\u1ebfp. Bi\u1ebft \u2206X, \u2206Y, \u2206Z l\u00e0 sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i t\u01b0\u01a1ng \u1ee9ng c\u1ee7a c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng X, Y, Z v\u00e0 F = X + Y \u2013 Z. H\u1ec7 th\u1ee9c \u0111\u00fang l\u00e0:<\/p>\n<\/div>\n

\n

A.<\/strong>\u2206F = \u2206X + \u2206Y – \u2206Z         <\/p>\n

B.<\/strong>\u2206F = (\u2206X + \u2206Y).\u2206Z<\/p>\n

C.<\/strong>\u2206F = (\u2206X + \u2206Y)\/\u2206Z         <\/p>\n

D.<\/strong>\u2206F = \u2206X + \u2206Y + \u2206Z<\/p>\n<\/div>\n

\n

C\u00e2u 18<\/strong>(ID: 82594)<\/strong>:<\/strong>Hai dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng t\u1ea7n s\u1ed1 v\u00e0 c\u00f9ng bi\u00ean \u0111\u1ed9 A. Dao \u0111\u1ed9ng t\u1ed5ng h\u1ee3p c\u1ee7a hai dao \u0111\u1ed9ng n\u00e0y c\u00f3 bi\u00ean \u0111\u1ed9 b\u1eb1ng 2A khi dao \u0111\u1ed9ng \u0111\u00f3<\/p>\n<\/div>\n

\n

A.<\/strong>l\u1ec7ch pha \u03c0\/3    <\/p>\n

B.<\/strong>c\u00f9ng pha    <\/p>\n

C.<\/strong>l\u1ec7ch pha 2\u03c0\/3     <\/p>\n

D.<\/strong>ng\u01b0\u1ee3c pha<\/p>\n<\/div>\n

\n

C\u00e2u 19<\/strong>(ID: 82595)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p u = U<\/p>\n

0<\/dd>\n

cos\u03c9t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L v\u00e0 t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C. Bi\u1ebft \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng gi\u1eefa hai \u0111\u1ea7u \u0111i\u1ec7n tr\u1edf thu\u1ea7n UR, gi\u1eefa hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m thu\u1ea7n U<\/p>\n

L<\/dd>\n

v\u00e0 gi\u1eefa hai \u0111\u1ea7u t\u1ee5 \u0111i\u1ec7n U<\/p>\n

C<\/dd>\n

tho\u1ea3 m\u00e3n UL = 2U<\/p>\n

R<\/dd>\n

= 2U<\/p>\n

C<\/dd>\n

. So v\u1edbi \u0111i\u1ec7n \u00e1p u, c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch<\/p>\n<\/div>\n

\n

A.<\/strong>tr\u1ec5 pha \u03c0\/4      <\/p>\n

B.<\/strong>tr\u1ec5 pha \u03c0\/2    <\/p>\n

C.<\/strong>s\u1edbm pha \u03c0\/4      <\/strong><\/p>\n

D.<\/strong>s\u1edbm pha \u03c0\/2<\/p>\n<\/div>\n

\n

C\u00e2u 20<\/strong>(ID: 82596)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p u = U\u221a2cos\u03c9t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L v\u00e0 t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong \u0111o\u1ea1n m\u1ea1ch l\u00e0 i = I\u221a2cos(\u03c9t \u2013 \u03c6). Khi \u0111\u00f3 \u0111o\u1ea1n m\u1ea1ch ti\u00eau th\u1ee5 c\u00f4ng su\u1ea5t b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>RIcos\u03c6       <\/p>\n

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

2<\/dd>\n

        <\/p>\n

C<\/strong>. UIcos\u03c6      <\/p>\n

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

\n

C\u00e2u 21<\/strong>(ID: 82597)<\/strong>:<\/strong>M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 th\u00ec b\u1eaft \u0111\u1ea7u ch\u1ecbu t\u00e1c d\u1ee5ng c\u1ee7a m\u1ed9t l\u1ef1c c\u1ea3n c\u00f3 \u0111\u1ed9 l\u1edbn kh\u00f4ng \u0111\u1ed5i th\u00ec v\u1eadt s\u1ebd<\/p>\n

A.<\/strong>dao \u0111\u1ed9ng \u1edf tr\u1ea1ng th\u00e1i c\u1ed9ng h\u01b0\u1edfng<\/p>\n

B.<\/strong>chuy\u1ec3n ngay sang th\u1ef1c hi\u1ec7n m\u1ed9t dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 v\u1edbi chu k\u00ec m\u1edbi<\/p>\n

C.<\/strong>dao \u0111\u1ed9ng v\u1edbi chu k\u00ec m\u1edbi sau th\u1eddi gian \u0111\u1ee7 l\u00e2u<\/p>\n

D.<\/strong>b\u1eaft \u0111\u1ea7u dao \u0111\u1ed9ng t\u1eaft d\u1ea7n<\/p>\n

C\u00e2u 22<\/strong>(ID: 82598)<\/strong>:<\/strong>Khi n\u00f3i v\u1ec1 s\u00f3ng \u00e2m, ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y kh\u00f4ng \u0111\u00fang?<\/p>\n

C\u00e2u 23<\/strong>(ID: 82599)<\/strong>:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng. K\u00edch th\u00edch cho con l\u1eafc dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 theo ph\u01b0\u01a1ng th\u1eb3ng \u0111\u1ee9ng. Ch\u1ecdn tr\u1ee5c x\u2019x th\u1eb3ng \u0111\u1ee9ng, chi\u1ec1u d\u01b0\u01a1ng h\u01b0\u1edbng xu\u1ed1ng d\u01b0\u1edbi, g\u1ed1c to\u1ea1 \u0111\u1ed9 t\u1ea1i v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng, gia t\u1ed1c r\u01a1i t\u1ef1 do g. T\u1ea7n s\u1ed1 v\u00e0 bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng c\u1ee7a con l\u1eafc l\u1ea7n l\u01b0\u1ee3t l\u00e0 f v\u00e0 g\/(2\u03c0<\/p>\n

2<\/dd>\n

f<\/p>\n

2<\/dd>\n

). Th\u1eddi gian ng\u1eafn nh\u1ea5t k\u1ec3 t\u1eeb khi l\u1ef1c \u0111\u00e0n h\u1ed3i c\u1ee7a l\u00f2 xo c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c ti\u1ec3u \u0111\u1ebfn khi l\u1ef1c \u0111\u00e0n h\u1ed3i c\u1ee7a l\u00f2 xo c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c \u0111\u1ea1i l\u00e0<\/p>\n<\/div>\n

\n

A<\/strong>. \u03c0f\/3      <\/p>\n

B<\/strong>. 1\/6f     <\/p>\n

C.<\/strong>\u03c0f\/6      <\/p>\n

D<\/strong>. 1\/3f<\/p>\n<\/div>\n

\n

C\u00e2u 24<\/strong>(ID: 82600)<\/strong>:<\/strong>H\u00ecnh v\u1ebd l\u00e0 \u0111\u1ed3 th\u1ecb bi\u1ec3u di\u1ec5n s\u1ef1 ph\u1ee5 thu\u1ed9c c\u1ee7a t\u1eeb th\u00f4ng qua m\u1ed9t v\u00f2ng d\u00e2y d\u1eabn. N\u1ebfu cu\u1ed9n d\u00e2y c\u00f3 200 v\u00f2ng d\u00e2y d\u1eabn th\u00ec bi\u1ec3u th\u1ee9c c\u1ee7a su\u1ea5t \u0111i\u1ec7n \u0111\u1ed9ng t\u1ea1o ra b\u1edfi cu\u1ed9n d\u00e2y l\u00e0: <\/p>\n<\/div>\n

\n

A.<\/strong>e = 251,2sin(20\u03c0t + 0,5\u03c0)          <\/p>\n

B.<\/strong>e = 251,2cos(20\u03c0t + 0,5\u03c0)<\/p>\n

C.<\/strong>e = 200cos(20\u03c0t + 0,5\u03c0)          <\/p>\n

D.<\/strong>e = 200sin20\u03c0t<\/p>\n<\/div>\n

\n

C\u00e2u 25<\/strong>(ID: 82601)<\/strong>:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng g\u1ed3m l\u00f2 xo nh\u1eb9, \u0111\u1ed9 c\u1ee9ng l\u00e0 50 N\/m v\u00e0 v\u1eadt n\u1eb7ng kh\u1ed1i l\u01b0\u1ee3ng 200g. K\u00e9o v\u1eadt th\u1eb3ng \u0111\u1ee9ng xu\u1ed1ng d\u01b0\u1edbi \u0111\u1ec3 l\u00f2 xo d\u00e3n 12 cm r\u1ed3i th\u1ea3 nh\u1eb9 cho v\u1eadt dao \u0111\u1ed9ng. B\u1ecf qua m\u1ecdi l\u1ef1c c\u1ea3n. L\u1ea5y g = 10m\/s<\/p>\n

2<\/dd>\n

v\u00e0 \u03c0<\/p>\n

2<\/dd>\n

= 10. Kho\u1ea3ng th\u1eddi gian l\u1ef1c \u0111\u00e0n h\u1ed3i t\u00e1c d\u1ee5ng v\u00e0o gi\u00e1 treo c\u00f9ng chi\u1ec1u v\u1edbi l\u1ef1c h\u1ed3i ph\u1ee5c trong m\u1ed9t chu k\u00ec l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>1\/30 s       <\/p>\n

B.<\/strong>4\/15 s      <\/p>\n

C.<\/strong>1\/3 s      <\/p>\n

D.<\/strong>1\/15 s<\/p>\n<\/div>\n

\n

C\u00e2u 26<\/strong>(ID: 82602)<\/strong>:<\/strong>Dao \u0111\u1ed9ng c\u01b0\u1ee1ng b\u1ee9c c\u1ee7a m\u1ed9t v\u1eadt do t\u00e1c d\u1ee5ng c\u1ee7a m\u1ed9t ngo\u1ea1i l\u1ef1c bi\u1ebfn thi\u00ean \u0111i\u1ec1u ho\u00e0 v\u1edbi t\u1ea7n s\u1ed1 f l\u00e0 dao \u0111\u1ed9ng c\u00f3 t\u1ea7n s\u1ed1<\/p>\n<\/div>\n

\n

A.<\/strong>2f       <\/p>\n

B.<\/strong>4f          <\/p>\n

C.<\/strong>f        <\/p>\n

D<\/strong>. 0,5f<\/p>\n<\/div>\n

\n

C\u00e2u 27<\/strong>(ID: 82603)<\/strong>:<\/strong>\u0110i\u1ec7n n\u0103ng \u0111\u01b0\u1ee3c truy\u1ec1n \u0111i v\u1edbi c\u00f4ng su\u1ea5t p tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng d\u00e2y t\u1ea3i \u0111i\u1ec7n d\u01b0\u1edbi \u0111i\u1ec7n \u00e1p U, hi\u1ec7u su\u1ea5t c\u1ee7a q\u00faa tr\u00ecnh truy\u1ec1n t\u1ea3i l\u00e0 n. Gi\u1eef nguy\u00ean \u0111i\u1ec7n \u00e1p tr\u00ean \u0111\u01b0\u1eddng d\u00e2y t\u1ea3i \u0111i\u1ec7n nh\u01b0ng t\u0103ng c\u00f4ng su\u1ea5t truy\u1ec1n t\u1ea3i l\u00ean k l\u1ea7n th\u00ec c\u00f4ng su\u1ea5t hao ph\u00ed tr\u00ean \u0111\u01b0\u1eddng d\u00e2y t\u1ea3i \u0111i\u1ec7n khi \u0111\u00f3 l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>  (1 \u2013 n)kp      <\/p>\n

B.<\/strong>(1+n)\/k<\/p>\n

2<\/dd>\n

p     <\/p>\n

C.<\/strong>(1-n)\/kp   <\/p>\n

D.<\/strong>(1 \u2013 n)k<\/p>\n

2<\/dd>\n

p<\/p>\n<\/div>\n

\n

C\u00e2u 28<\/strong>(ID: 82604)<\/strong>:<\/strong>\u0110\u1ec3 kh\u1eafc ph\u1ee5c t\u00ecnh tr\u1ea1ng \u0111ua xe moto tr\u00ean \u0111\u01b0\u1eddng ph\u1ed1, \u1edf g\u1ea7n c\u00e1c khu d\u00e2n c\u01b0, ng\u01b0\u1eddi ta th\u01b0\u1eddng l\u00e0m c\u00e1c d\u1ea3i song song \u0111\u1ec3 t\u1ea1o th\u00e0nh c\u00e1c g\u1edd l\u00e0m gi\u1ea3m t\u1ed1c \u0111\u1ed9 c\u1ee7a xe khi \u0111i qua v\u00e0 g\u1ecdi l\u00e0 g\u1edd gi\u1ea3m t\u1ed1c. Khi xe moto \u0111i qua v\u1edbi t\u1ed1c \u0111\u1ed9 45 km\/h th\u00ec xe b\u1ecb x\u00f3c m\u1ea1nh nh\u1ea5t v\u00e0 l\u00f2 xo gi\u1ea3m x\u00f3c c\u1ee7a xe l\u00fac \u0111\u00f3 dao \u0111\u1ed9ng v\u1edbi t\u1ea7n s\u1ed1 50 Hz. Kho\u1ea3ng c\u00e1ch gi\u1eefa hai g\u1edd song song li\u00ean ti\u1ebfp b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>20 cm      <\/p>\n

B.<\/strong>25 cm       <\/p>\n

C.<\/strong>50 cm      <\/p>\n

D.<\/strong>45 cm<\/p>\n<\/div>\n

\n

C\u00e2u 29<\/strong>(ID: 82605)<\/strong>:<\/strong>Cho \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n xoay chi\u1ec1u AB m\u1eafc n\u1ed1i ti\u1ebfp theo th\u1ee9 t\u1ef1 g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C v\u00e0 cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L thay \u0111\u1ed5i \u0111\u01b0\u1ee3c. \u0110\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n \u00e1p u = 90\u221a10cos\u03c9t (V) (\u03c9 kh\u00f4ng \u0111\u1ed5i). Khi Z<\/p>\n

L<\/dd>\n

= Z<\/p>\n

L1<\/dd>\n

ho\u1eb7c Z<\/p>\n

L<\/dd>\n

= Z<\/p>\n

L2<\/dd>\n

th\u00ec \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m c\u00f3 c\u00f9ng gi\u00e1 tr\u1ecb hi\u1ec7u d\u1ee5ng U<\/p>\n

L<\/dd>\n

= 270 V. Bi\u1ebft r\u1eb1ng 3Z<\/p>\n

L2<\/dd>\n

\u2013 Z<\/p>\n

L1<\/dd>\n

= 150 \u2126 v\u00e0 t\u1ed5ng tr\u1edf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch RC trong hai tr\u01b0\u1eddng h\u1ee3p \u0111\u1ec1u l\u00e0 100\u221a2 \u2126 . \u0110\u1ec3 \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i th\u00ec c\u1ea3m kh\u00e1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB khi \u0111\u00f3 g\u1ea7n gi\u00e1 tr\u1ecb n\u00e0o nh\u1ea5t?<\/p>\n<\/div>\n

\n

A<\/strong>. 150 \u2126      <\/p>\n

B.<\/strong>180 \u2126      <\/p>\n

C.<\/strong>175 \u2126         <\/p>\n

D.<\/strong>192 \u2126<\/p>\n<\/div>\n

\n

C\u00e2u 30<\/strong>(ID: 82606)<\/strong>:<\/strong>\u0110i\u1ec7n \u00e1p gi\u1eefa hai \u0111\u1ea7u m\u1ed9t \u0111o\u1ea1n m\u1ea1ch c\u00f3 bi\u1ec3u th\u1ee9c u = 220cos100\u03c0t (V). Gi\u00e1 tr\u1ecb hi\u1ec7u d\u1ee5ng c\u1ee7a \u0111i\u1ec7n \u00e1p n\u00e0y l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>  110V       <\/p>\n

B.<\/strong>110\u221a2 V      <\/p>\n

C.<\/strong>220V       <\/p>\n

D.<\/strong>220\u221a2 V<\/p>\n<\/div>\n

\n

C\u00e2u 31<\/strong>(ID: 82607)<\/strong>:<\/strong>M\u1ed9t m\u1ea1ch dao \u0111\u1ed9ng \u0111i\u1ec7n t\u1eeb LC c\u00f3 chu k\u00ec dao \u0111\u1ed9ng ri\u00eang l\u00e0 T. T\u1ea1i th\u1eddi \u0111i\u1ec3m ban \u0111\u1ea7u, \u0111i\u1ec7n t\u00edch tr\u00ean t\u1ee5 b\u1eb1ng kh\u00f4ng. Sau \u0111\u00f3 m\u1ed9t kho\u1ea3ng th\u1eddi gian ng\u1eafn nh\u1ea5t l\u00e0 bao nhi\u00eau th\u00ec \u0111i\u1ec7n t\u00edch c\u1ee7a t\u1ee5 c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng m\u1ed9t n\u1eeda gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i<\/p>\n<\/div>\n

\n

A.<\/strong>T\/12      <\/p>\n

B.<\/strong>T\/6        <\/p>\n

C.<\/strong>T\/3         <\/p>\n

D.<\/strong>T\/4<\/p>\n<\/div>\n

\n

C\u00e2u 32<\/strong>(ID: 82608)<\/strong>:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng t\u1eaft d\u1ea7n tr\u00ean tr\u1ee5c Ox do c\u00f3 ma s\u00e1t gi\u1eefa v\u1eadt v\u00e0 m\u1eb7t ph\u1eb3ng n\u1eb1m ngang. C\u1ee9 sau m\u1ed7i chu k\u00ec dao \u0111\u1ed9ng, bi\u00ean \u0111\u1ed9 c\u1ee7a v\u1eadt l\u1ea1i gi\u1ea3m \u0111i 2%. So v\u1edbi c\u01a1 n\u0103ng ban \u0111\u1ea7u th\u00ec ph\u1ea7n c\u01a1 n\u0103ng c\u00f2n l\u1ea1i sau 5 chu k\u00ec dao \u0111\u1ed9ng b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>81%       <\/p>\n

B.<\/strong>91%       <\/p>\n

C.<\/strong>19%         <\/p>\n

D.<\/strong>9%<\/p>\n<\/div>\n

\n

C\u00e2u 33<\/strong>(ID: 82609)<\/strong>:<\/strong>Trong dao \u0111\u1ed9ng c\u01b0\u1ee1ng b\u1ee9c c\u1ee7a m\u1ed9t v\u1eadt, khi x\u1ea3y ra c\u1ed9ng h\u01b0\u1edfng th\u00ec v\u1eadt ti\u1ebfp t\u1ee5c dao \u0111\u1ed9ng<\/p>\n

A<\/strong>. v\u1edbi t\u1ea7n s\u1ed1 b\u1eb1ng t\u1ea7n s\u1ed1 dao \u0111\u1ed9ng ri\u00eang c\u1ee7a h\u1ec7<\/p>\n

B.<\/strong>  v\u1edbi t\u1ea7n s\u1ed1 nh\u1ecf h\u01a1n t\u1ea7n s\u1ed1 dao \u0111\u1ed9ng ri\u00eang c\u1ee7a h\u1ec7  <\/p>\n

C.<\/strong>v\u1edbi t\u1ea7n s\u1ed1 l\u1edbn h\u01a1n t\u1ea7n s\u1ed1 dao \u0111\u1ed9ng ri\u00eang c\u1ee7a h\u1ec7 <\/p>\n

D.<\/strong>m\u00e0 kh\u00f4ng ch\u1ecbu ngo\u1ea1i l\u1ef1c t\u00e1c d\u1ee5ng<\/p>\n

C\u00e2u 34<\/strong>(ID: 82610)<\/strong>:<\/strong>M\u1ed9t \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n xoay chi\u1ec1u g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi t\u1ee5 \u0111i\u1ec7n C. N\u1ebfu dung kh\u00e1ng b\u1eb1ng R th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch<\/p>\n

A.<\/strong>nhanh pha \u03c0\/4 so v\u1edbi \u0111i\u1ec7n \u00e1p \u1edf hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch<\/p>\n

B.<\/strong>ch\u1eadm pha \u03c0\/2 so v\u1edbi \u0111i\u1ec7n \u00e1p \u1edf hai \u0111\u1ea7u t\u1ee5 \u0111i\u1ec7n<\/p>\n

C.<\/strong>nhanh pha \u03c0\/2 so v\u1edbi \u0111i\u1ec7n \u00e1p \u1edf hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch<\/p>\n

D<\/strong>. ch\u1eadm pha \u03c0\/4 so v\u1edbi \u0111i\u1ec7n \u00e1p \u1edf hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch<\/p>\n

C\u00e2u 35<\/strong>(ID: 82611)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p xoay chi\u1ec1u u  = U<\/p>\n

0<\/dd>\n

cos2\u03c0ft (trong \u0111\u00f3 U<\/p>\n

0<\/dd>\n

kh\u00f4ng \u0111\u1ed5i, f thay \u0111\u1ed5i \u0111\u01b0\u1ee3c) v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n, cu\u1ed9n c\u1ea3m thu\u1ea7n v\u00e0 t\u1ee5 \u0111i\u1ec7n. Ban \u0111\u1ea7u trong \u0111o\u1ea1n m\u1ea1ch \u0111ang c\u00f3 c\u1ed9ng h\u01b0\u1edfng \u0111i\u1ec7n. Gi\u1ea3m t\u1ea7n s\u1ed1 f th\u00ec \u0111i\u1ec7n \u00e1p u s\u1ebd<\/p>\n<\/div>\n

\n

A.<\/strong>tr\u1ec5 pha so v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n        <\/p>\n

B.<\/strong>s\u1edbm pha so v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n        <\/p>\n

C.<\/strong>ng\u01b0\u1ee3c pha so v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n       <\/p>\n

D.<\/strong>c\u00f9ng pha so v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n<\/p>\n<\/div>\n

\n

C\u00e2u 36<\/strong>(ID: 82612)<\/strong>:<\/strong>\u0110\u1eb7t \u0111i\u1ec7n \u00e1p u = U<\/p>\n

0<\/dd>\n

cos100\u03c0t (V) (U<\/p>\n

0<\/dd>\n

kh\u00f4ng \u0111\u1ed5i) v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp theo th\u1ee9 t\u1ef1 g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, t\u1ee5 \u0111i\u1ec7n C v\u00e0 cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L thay \u0111\u1ed5i \u0111\u01b0\u1ee3c. Khi L = L<\/p>\n

0<\/dd>\n

th\u00ec \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng gi\u1eefa hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i l\u00e0 U<\/p>\n

L max<\/dd>\n

. Khi L = L<\/p>\n

1<\/dd>\n

ho\u1eb7c L = L<\/p>\n

2<\/dd>\n

th\u00ec \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng gi\u1eefa hai \u0111\u1ea7u cu\u1ed9n c\u1ea3m c\u00f3 gi\u00e1 tr\u1ecb nh\u01b0 nhau l\u00e0 U<\/p>\n

L<\/dd>\n

= kU<\/p>\n

L max<\/dd>\n

. G\u1ecdi cos\u03c6<\/p>\n

1<\/dd>\n

, cos\u03c6<\/p>\n

2<\/dd>\n

v\u00e0 cos\u03c6<\/p>\n

\u03b1<\/dd>\n

l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u1ec7 s\u1ed1 c\u00f4ng su\u1ea5t c\u1ee7a m\u1ea1ch khi \u0111\u1ed9 t\u1ef1 c\u1ea3m l\u00e0 L<\/p>\n

1<\/dd>\n

, L<\/p>\n

2<\/dd>\n

v\u00e0 L<\/p>\n

\u03b1<\/dd>\n

. Bi\u1ebft r\u1eb1ng cos\u03c6<\/p>\n

1<\/dd>\n

+ cos\u03c6<\/p>\n

2<\/dd>\n

= k\u221a2. Gi\u00e1 tr\u1ecb c\u1ee7a cos\u03c6\u03b1 b\u1eb1ng<\/p>\n<\/div>\n

\n

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

B.<\/strong>\u221a2\/2         <\/p>\n

C.<\/strong>\u00be         <\/p>\n

C.<\/strong>\u00bd<\/p>\n<\/div>\n

\n

C\u00e2u 37<\/strong>(ID: 82613)<\/strong>:<\/strong>Bi\u1ebft c\u01b0\u1eddng \u0111\u1ed9 \u00e2m chu\u1ea9n cuae m\u1ed9t \u00e2m l\u00e0 10<\/p>\n

-12<\/dd>\n

W\/m<\/p>\n

2<\/dd>\n

. M\u1ee9c c\u01b0\u1eddng \u0111\u1ed9 \u00e2m c\u1ee7a m\u1ed9t \u00e2m t\u1ea1i m\u1ed9t \u0111i\u1ec3m l\u00e0 50dB th\u00ec c\u01b0\u1eddng \u0111\u1ed9 c\u1ee7a \u00e2m t\u1ea1i \u0111i\u1ec3m \u0111\u00f3 b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>10W\/m<\/p>\n

2<\/dd>\n

       <\/p>\n

B.<\/strong>10<\/p>\n

-7<\/dd>\n

W\/m<\/p>\n

2<\/dd>\n

       <\/p>\n

C.<\/strong>50 W\/m<\/p>\n

2<\/dd>\n

       <\/p>\n

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

-12<\/dd>\n

W\/m<\/p>\n

2<\/dd>\n

 <\/p>\n<\/div>\n

\n

C\u00e2u 38<\/strong>(ID: 82614)<\/strong>:<\/strong>  H\u1ec7 th\u1ee9c x\u00e1c \u0111\u1ecbnh chu k\u00ec c\u1ee7a dao \u0111\u1ed9ng \u0111i\u1ec7n t\u1eeb t\u1ef1 do trong m\u1ea1ch dao \u0111\u1ed9ng LC c\u00f3 \u0111i\u1ec7n tr\u1edf thu\u1ea7n kh\u00f4ng \u0111\u00e1ng k\u1ec3 l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong> 2\u03c0\/\u221a(LC)       <\/p>\n

B.<\/strong>2\u03c0\/\u221a(L\/C)      <\/p>\n

C.<\/strong>2\u03c0\/\u221a(C\/L)     <\/p>\n

D.<\/strong>2\u03c0\u221a(LC)<\/p>\n<\/div>\n

\n

C\u00e2u 39<\/strong>(ID: 82615)<\/strong>:<\/strong>Ph\u00e1t bi\u1ec3u n\u00e0o l\u00e0 \u0111\u00fang khi n\u00f3i v\u1ec1 si\u00eau \u00e2m?<\/p>\n

A. <\/strong>Si\u00eau \u00e2m \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng ghi h\u00ecnh \u1ea3nh trong c\u01a1 th\u1ec3 ng\u01b0\u1eddi \u0111\u1ec3 chu\u1ea9n \u0111o\u00e1n b\u1ec7nh<\/p>\n

B.  <\/strong>Si\u00eau \u00e2m c\u00f3 b\u01b0\u1edbc s\u00f3ng l\u1edbn n\u00ean tai ng\u01b0\u1eddi kh\u00f4ng nghe  \u0111\u01b0\u1ee3c si\u00eau \u00e2m<\/p>\n

C. <\/strong>Si\u00eau \u00e2m truy\u1ec1n \u0111\u01b0\u1ee3c qua c\u00e1c v\u1eadt r\u1eafn v\u00e0 kh\u00f4ng ph\u1ea3n x\u1ea1 \u1edf m\u1eb7t ti\u1ebfp x\u00fac gi\u1eefa hai v\u1eadt<\/p>\n

D. <\/strong>Si\u00eau \u00e2m l\u00e0 s\u00f3ng c\u01a1 h\u1ecdc c\u00f3 t\u1ea7n s\u1ed1 nh\u1ecf h\u01a1n 20 kHz<\/p>\n

C\u00e2u 40<\/strong>(ID: 82616)<\/strong>:<\/strong>T\u1ea1i m\u1eb7t n\u01b0\u1edbc c\u00f3 hai ngu\u1ed3n s\u00f3ng k\u1ebft h\u1ee3p S<\/p>\n

1<\/dd>\n

, S<\/p>\n

2<\/dd>\n

c\u00e1ch nhau 12 cm, dao \u0111\u1ed9ng \u0111\u1ed3ng pha nhau v\u1edbi t\u1ea7n s\u1ed1 20 Hz. \u0110i\u1ec3m M c\u00e1ch S<\/p>\n

1<\/dd>\n

, S<\/p>\n

2<\/dd>\n

l\u1ea7n l\u01b0\u1ee3t 4,2 cm v\u00e0 9 cm. Bi\u1ebft t\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng tr\u00ean m\u1eb7t n\u01b0\u1edbc l\u00e0 32 cm\/s. \u0110\u1ec3 M thu\u1ed9c v\u00e2n c\u1ef1c ti\u1ec3u th\u00ec ph\u1ea3i d\u1ecbch chuy\u1ec3n S<\/p>\n

2<\/dd>\n

theo ph\u01b0\u01a1ng S<\/p>\n

1<\/dd>\n

S<\/p>\n

2<\/dd>\n

ra xaS1 m\u1ed9t kho\u1ea3ng t\u1ed1i thi\u1ec3u l\u00e0<\/p>\n<\/div>\n

\n

A. <\/strong>4,80 cm        <\/p>\n

B.<\/strong>1,62 cm       <\/p>\n

C.<\/strong>0,83 cm      <\/p>\n

D.<\/strong>0,54 cm<\/p>\n<\/div>\n

\n

C\u00e2u 41<\/strong>(ID: 82617)<\/strong>:<\/strong>M\u1ed9t v\u1eadt kh\u1ed1i l\u01b0\u1ee3ng m, th\u1ef1c hi\u1ec7n dao \u0111\u1ed9ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x<\/p>\n

1<\/dd>\n

= A<\/p>\n

1<\/dd>\n

cos(2\u03c0t + \u03c0\/3) cm th\u00ec c\u01a1 n\u0103ng l\u00e0 W<\/p>\n

1<\/dd>\n

. N\u1ebfu cho v\u1eadt dao \u0111\u1ed9ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x<\/p>\n

2<\/dd>\n

= A<\/p>\n

2<\/dd>\n

cos(2\u03c0t ) cm th\u00ec c\u01a1 n\u0103ng l\u00e0 W<\/p>\n

2<\/dd>\n

= 4W<\/p>\n

1<\/dd>\n

. Bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa A<\/p>\n

2<\/dd>\n

v\u00e0 A<\/p>\n

1<\/dd>\n

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

\n

A. <\/strong>A2 = A1       <\/p>\n

B.<\/strong>A2 = 2A1         <\/p>\n

C.<\/strong>A2 = 4A1      <\/p>\n

D.<\/strong>A2 = 0,5A1     <\/p>\n<\/div>\n

\n

C\u00e2u 42<\/strong>(ID: 82618)<\/strong>:<\/strong>D\u00f2ng \u0111i\u1ec7n t\u1ee9c th\u1eddi trong m\u1ea1ch dao \u0111\u1ed9ng \u0111i\u1ec7n t\u1eeb LC l\u00e0 i = I<\/p>\n

0<\/dd>\n

sin2000t (I<\/p>\n

0<\/dd>\n

kh\u00f4ng \u0111\u1ed5i, t t\u00ednh b\u1eb1ng gi\u00e2y). T\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C = 5\u03bcF. \u0110\u1ed9 t\u1ef1 c\u1ea3m L c\u1ee7a cu\u1ed9n d\u00e2y thu\u1ea7n c\u1ea3m l\u00e0<\/p>\n<\/div>\n

\n

A. 25mH         <\/p>\n

B.<\/strong>100 mH<\/p>\n

C.<\/strong>50 mH       <\/p>\n

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

\n

C\u00e2u 43<\/strong>(ID: 82619)<\/strong>:<\/strong>Trong m\u1ed9t m\u00f4i tr\u01b0\u1eddng c\u00f3 s\u00f3ng t\u1ea7n s\u1ed1 50 Hz, lan truy\u1ec1n v\u1edbi t\u1ed1c \u0111\u1ed9 160 m\/s. Hai \u0111i\u1ec3m g\u1ea7n nhau nh\u1ea5t tr\u00ean c\u00f9ng m\u1ed9t ph\u01b0\u01a1ng truy\u1ec1n s\u00f3ng dao \u0111\u1ed9ng l\u1ec7ch pha nhau \u03c0\/4, c\u00e1ch nhau<\/p>\n<\/div>\n

\n

A.<\/strong>0,8 m        <\/p>\n

B.<\/strong>1,6 cm          <\/p>\n

C.<\/strong>0,4 m        <\/p>\n

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

\n

C\u00e2u 44<\/strong>(ID: 82620)<\/strong>:<\/strong>M\u1ed9t \u0111o\u1ea1n m\u1ea1ch xoay chi\u1ec1u m\u1eafc n\u1ed1i ti\u1ebfp g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L v\u00e0 t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C. Trong \u0111o\u1ea1n m\u1ea1ch \u0111ang c\u00f3 c\u1ed9ng h\u01b0\u1edfng \u0111i\u1ec7n, ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y sai?<\/p>\n

A. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong \u0111o\u1ea1n m\u1ea1ch s\u1edbm pha so v\u1edbi \u0111i\u1ec7n \u00e1p gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch.<\/p>\n

B. H\u1ec7 s\u1ed1 c\u00f4ng su\u1ea5t c\u1ee7a \u0111o\u1ea1n m\u1ea1ch b\u1eb1ng 1<\/p>\n

C. \u0110i\u1ec7n \u00e1p gi\u1eefa hai \u0111\u1ea7u \u0111i\u1ec7n tr\u1edf thu\u1ea7n c\u00f9ng pha v\u1edbi \u0111i\u1ec7n \u00e1p gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch<\/p>\n

D. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n hi\u1ec7u d\u1ee5ng trong m\u1ea1ch c\u1ef1c \u0111\u1ea1i<\/p>\n

C\u00e2u 45<\/strong>(ID: 82621)<\/strong>:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0 d\u1ecdc tr\u00ean tr\u1ee5c Ox theo ph\u01b0\u01a1ng ngang, xung quanh v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng O, v\u1edbi bi\u00ean \u0111\u1ed9 l\u00e0 x<\/p>\n

m<\/dd>\n

. Ch\u1ecdn m\u1ed1c th\u1ebf n\u0103ng \u0111\u00e0n h\u1ed3i tr\u00f9ng v\u1edbi g\u1ed1c to\u1ea1 \u0111\u1ed9 O. Khi v\u1eadt c\u00f3 \u0111\u1ed9ng n\u0103ng b\u1eb1ng th\u1ebf n\u0103ng th\u00ec li \u0111\u1ed9 c\u1ee7a v\u1eadt l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>x = \u00b1  (x<\/p>\n

m<\/dd>\n

\u221a2)\/4    <\/p>\n

B.<\/strong>x = \u00b1  (x<\/p>\n

m<\/dd>\n

\u221a2)\/2       <\/p>\n

C.<\/strong>x = \u00b1  x<\/p>\n

m<\/dd>\n

\/2   <\/p>\n

D.<\/strong>   x = \u00b1  x<\/p>\n

m<\/dd>\n

\/4       <\/p>\n<\/div>\n

\n

 C\u00e2u 46<\/strong>(ID: 82622)<\/strong>:<\/strong>M\u1ea1ch \u0111i\u1ec7n AB g\u1ed3m \u0111o\u1ea1n m\u1ea1ch AM n\u1ed1i ti\u1ebfp v\u1edbi \u0111o\u1ea1n m\u1ea1ch MB. \u0110\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n \u00e1p u = 150\u221a2cos100\u03c0t (V). \u0110i\u1ec7n \u00e1p \u1edf hai \u0111\u1ea7u \u0111o\u1ea1n AM s\u1edbm pha h\u01a1n c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n m\u1ed9t g\u00f3c \u03c0\/6. \u0110o\u1ea1n MB ch\u1ec9 c\u00f3 t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C thay \u0111\u1ed5i \u0111\u01b0\u1ee3c. \u0110i\u1ec1u ch\u1ec9nh C \u0111\u1ec3 t\u1ed5ng \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng c\u1ee7a \u0111o\u1ea1n AM v\u00e0 \u0111o\u1ea1n MB \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i. Khi \u0111\u00f3 \u0111i\u1ec7n \u00e1p hi\u1ec7u d\u1ee5ng \u1edf hai \u0111\u1ea7u t\u1ee5 \u0111i\u1ec7n l\u00e0<\/p>\n<\/div>\n

\n

A.<\/strong>75\u221a3 V       <\/p>\n

B.<\/strong>150V        <\/p>\n

C.<\/strong>75\u221a2 V          <\/p>\n

D.<\/strong>200V<\/p>\n<\/div>\n

\n

C\u00e2u 47<\/strong>(ID: 82623)<\/strong>: Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang khi n\u00f3i v\u1ec1 s\u00f3ng c\u01a1 h\u1ecdc?<\/p>\n

C\u00e2u 48<\/strong>(ID: 82624)<\/strong>: \u0110\u1eb7t \u0111i\u1ec7n \u00e1p u = U\u221a2cos\u03c9t (U kh\u00f4ng \u0111\u1ed5i, \u03c9 thay \u0111\u1ed5i \u0111\u01b0\u1ee3c) v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m \u0111i\u1ec7n tr\u1edf thu\u1ea7n R, cu\u1ed9n c\u1ea3m thu\u1ea7n c\u00f3 \u0111\u1ed9 t\u1ef1 c\u1ea3m L = 2,5\/\u03c0 (H) v\u00e0 t\u1ee5 \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n dung C m\u1eafc n\u1ed1i ti\u1ebfp. Thay \u0111\u1ed5i t\u1ea7n s\u1ed1 g\u00f3c \u03c9 th\u00ec th\u1ea5y khi \u03c9 = 60\u03c0 (rad\/s), c\u01b0\u1eddng \u0111\u1ed9 hi\u1ec7u d\u1ee5ng d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch l\u00e0 I<\/p>\n

1<\/dd>\n

. Khi \u03c9 = 40\u03c0 (rad\/s), c\u01b0\u1eddng \u0111\u1ed9 hi\u1ec7u d\u1ee5ng d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch l\u00e0 I<\/p>\n

2<\/dd>\n

. Khi t\u1ea7n s\u1ed1 l\u00e0  \u03c9 = \u03c90 th\u00ec c\u01b0\u1eddng \u0111\u1ed9 hi\u1ec7u d\u1ee5ng c\u1ee7a d\u00f2ng \u0111i\u1ec7n \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i I max v\u00e0 I<\/p>\n

1<\/dd>\n

= I<\/p>\n

2<\/dd>\n

= I<\/p>\n

max<\/dd>\n

\/\u221a5 . Gi\u00e1 tr\u1ecb c\u1ee7a R b\u1eb1ng<\/p>\n<\/div>\n

\n

A.<\/strong>75 \u2126        <\/p>\n

B.<\/strong>100 \u2126        <\/p>\n

C.<\/strong>50 \u2126       <\/p>\n

D<\/strong>. 12,5\u2126<\/p>\n<\/div>\n

\n

C\u00e2u 49<\/strong>(ID: 82625)<\/strong>:<\/strong>M\u1ed9t con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng \u0111i\u1ec1u ho\u00e0. N\u1ebfu ch\u1ec9 thay \u0111\u1ed5i c\u00e1ch ch\u1ecdn g\u1ed1c th\u1eddi gian th\u00ec<\/p>\n

A.<\/strong>Bi\u00ean \u0111\u1ed9 v\u00e0 chu k\u00ec thay \u0111\u1ed5i c\u00f2n pha dao \u0111\u1ed9ng ban \u0111\u1ea7u kh\u00f4ng \u0111\u1ed5i<\/p>\n

B.<\/strong>C\u1ea3 bi\u00ean \u0111\u1ed9, chu k\u00ec v\u00e0 pha ban \u0111\u1ea7u c\u1ee7a dao \u0111\u1ed9ng \u0111\u1ec1u thay \u0111\u1ed5i<\/p>\n

C.<\/strong>C\u1ea3 bi\u00ean \u0111\u1ed9, chu k\u00ec v\u00e0 pha ban \u0111\u1ea7u c\u1ee7a dao \u0111\u1ed9ng \u0111\u1ec1u kh\u00f4ng thay \u0111\u1ed5i<\/p>\n

D.<\/strong>Bi\u00ean \u0111\u1ed9 v\u00e0 chu k\u00ec kh\u00f4ng \u0111\u1ed5i c\u00f2n pha ban \u0111\u1ea7u c\u1ee7a dao \u0111\u1ed9ng thay \u0111\u1ed5i<\/p>\n

C\u00e2u 50<\/strong>(ID: 82626)<\/strong>:<\/strong>Tr\u00ean m\u1ed9t s\u1ee3i d\u00e2y \u0111\u00e0n h\u1ed3i \u0111ang c\u00f3 s\u00f3ng d\u1eebng v\u1edbi t\u1ea7n s\u1ed1 100Hz, ng\u01b0\u1eddi ta th\u1ea5y kho\u1ea3ng c\u00e1ch gi\u1eefa hai n\u00fat s\u00f3ng li\u00ean ti\u1ebfp l\u00e0 0,5m. T\u1ed1c \u0111\u1ed9 truy\u1ec1n s\u00f3ng tr\u00ean d\u00e2y l\u00e0:<\/p>\n<\/div>\n

\n

A.<\/strong>40 m\/s      <\/p>\n

B.<\/strong>60 m\/s        <\/p>\n

C.<\/strong>80 m\/s        <\/p>\n

D.<\/strong>100 m\/s<\/p>\n

\u0110\u00e1p \u00e1n \u0111\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd n\u0103m 2016 – \u0110\u1ec1 s\u1ed1 1<\/strong><\/span><\/h3>\n

C\u00e2u 1<\/u><\/strong><\/p>\n

 \u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 2:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A. <\/p>\n

C\u00e2u 3:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A. <\/p>\n

C\u00e2u 4:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 5:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D. <\/p>\n

Cau 6:<\/u><\/strong>\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 7:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A.<\/strong><\/p>\n

C\u00e2u 8:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D. <\/p>\n

C\u00e2u 9<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A. <\/p>\n

C\u00e2u 10:<\/u><\/strong>\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 11:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A.<\/p>\n

C\u00e2u 12:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B<\/p>\n

C\u00e2u 13 :<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n C<\/p>\n

C\u00e2u 14:<\/u><\/strong>\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 15:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 16:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 17:<\/u><\/strong>\u0110\u00e1p an D.<\/strong><\/p>\n

C\u00e2u 18:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B.<\/p>\n

C\u00e2u 19:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A.<\/p>\n

C\u00e2u 20:<\/u><\/strong>\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 21:<\/u><\/strong>\u0110\u00e1p \u00e1n D. <\/p>\n

C\u00e2u 22:<\/u><\/strong>\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 23:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D.<\/p>\n

C\u00e2u 24:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 25:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D . <\/p>\n

C\u00e2u 26:<\/u><\/strong>\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 27:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D. <\/p>\n

C\u00e2u 28:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B.<\/p>\n

C\u00e2u 29:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D.<\/p>\n

C\u00e2u 30:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 31<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A. <\/p>\n

C\u00e2u 32:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A.<\/p>\n

C\u00e2u 33:<\/u><\/strong>\u0110\u00e1p \u00e1n A.<\/p>\n

C\u00e2u 34:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A.
C\u00e2u 35:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n A.<\/p>\n

 C\u00e2u 36:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B.<\/p>\n

C\u00e2u 37:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 38:<\/u><\/strong>\u0110\u00e1p \u00e1n D. <\/p>\n

C\u00e2u 39:<\/u><\/strong>\u0110\u00e1p \u00e1n A. <\/p>\n

C\u00e2u 40<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n C.<\/p>\n

C\u00e2u 41:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 42:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 43:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 44:<\/u><\/strong>\u0110\u00e1p \u00e1n A. <\/p>\n

C\u00e2u 45:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 46:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n B. <\/p>\n

C\u00e2u 47:<\/u><\/strong>\u0110\u00e1p \u00e1n C. <\/p>\n

C\u00e2u 48:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D <\/p>\n

C\u00e2u 49:<\/u><\/strong>\u0110\u00e1p \u00e1n D. <\/p>\n

C\u00e2u 50:<\/u><\/strong><\/p>\n

\u0110\u00e1p \u00e1n D.<\/p>\n<\/div>\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":"

C\u1eadp nh\u1eadt \u0111\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n v\u1eadt l\u00fd n\u0103m 2016 gi\u00fap c\u00e1c em l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 thi THPT Qu\u1ed1c gia s\u1eafp t\u1edbi. \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n L\u00fd n\u0103m 2016 – \u0110\u1ec1 s\u1ed1 1  C\u00e2u 1(ID: 82547):S\u00f3ng d\u1eebng xu\u1ea5t hi\u1ec7n tr\u00ean m\u1ed9t s\u1ee3i d\u00e2y […]<\/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