Bài 3: Thực hành tính sai số trong phép đo. Ghi kết quả đo

Home Lớp 10 Vật lý 10 - Sách kết nối tri thức Bài 3: Thực hành tính sai số trong phép đo. Ghi kết quả đoBài tập nâng cao

{"common":{"save":0,"post_id":"6287","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5656","mon_id":"1158790","chapter_id":"1158791","post_id":"6287","question":"Trong các phép tính d\u01b0\u1edbi \u0111ây, phép tính nào sai?<\/p>","options":["A. $\\triangle A = \\bar{\\triangle A}+\\triangle A_{dc}$<\/span>","B. $\\triangle A = \\delta A.\\bar{A}$<\/span><\/strong>","C. $\\bar{A}=\\frac{A_1+A_2+...+A_n}{n}$<\/span>","D. $\\bar{\\triangle A}=\\frac{A_1+A_2+...+A_n}{n}$<\/span>"],"correct":"4","level":"3","hint":"- Sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i c\u1ee7a phép \u0111o:   $\\triangle A = \\bar{\\triangle A}+\\triangle A_{dc}$<\/span> - Sai s\u1ed1 t\u1ec9 \u0111\u1ed1i c\u1ee7a phép \u0111o:   $\\delta A = \\frac{\\triangle A}{\\bar A}.100$<\/span>% - Giá tr\u1ecb trung bình c\u1ee7a m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng c\u1ea7n \u0111o:   $\\bar{A}=\\frac{A_1+A_2+...+A_n}{n}$<\/span> - Sai s\u1ed1 ng\u1eabu nhiên tuy\u1ec7t \u0111\u1ed1i trung bình c\u1ee7a n l\u1ea7n \u0111o:  $\\bar{ \\triangle A}=\\frac{\\triangle A_1+\\triangle A_2+...+\\triangle A_n}{n}$<\/span><\/p>","answer":"Ch\u1ecdn D. - Sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i c\u1ee7a phép \u0111o b\u1eb1ng t\u1ed5ng sai s\u1ed1 ng\u1eabu nhiên và sai s\u1ed1 d\u1ee5ng c\u1ee5:$\\triangle A = \\bar{\\triangle A}+\\triangle A_{dc}$<\/span>   => A \u0111úng. - Sai s\u1ed1 t\u1ec9 \u0111\u1ed1i c\u1ee7a phép \u0111o là t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m gi\u1eefa sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i và giá tr\u1ecb trung bình c\u1ee7a \u0111\u1ea1i l\u01b0\u1ee3ng c\u1ea7n \u0111o:<\/strong><\/span><\/p>$\\delta A = \\frac{\\triangle A}{\\bar A}.100$<\/span>%$\\to \\triangle A = \\delta A.\\bar{A}$<\/span>   => B \u0111úng. - Giá tr\u1ecb trung bình c\u1ee7a m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng c\u1ea7n \u0111o là phép tính trung bình c\u1ed9ng c\u1ee7a các k\u1ebft qu\u1ea3 \u0111o$\\bar{A}=\\frac{A_1+A_2+...+A_n}{n}$<\/span>:   => C \u0111úng. - Sai s\u1ed1 ng\u1eabu nhiên tuy\u1ec7t \u0111\u1ed1i trung bình c\u1ee7a n l\u1ea7n \u0111o \u0111\u01b0\u1ee3c tính theo công th\u1ee9c:$\\bar{ \\triangle A}=\\frac{\\triangle A_1+\\triangle A_2+...+\\triangle A_n}{n}$<\/span>   => D sai.<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-09-29 05:19:44","option_type":"math","len":0},{"id":"5657","mon_id":"1158790","chapter_id":"1158791","post_id":"6287","question":"Phép \u0111o \u0111\u1ed9 dài quãng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c s cho giá tr\u1ecb trung bình  $\\bar{s}=1,36832$<\/span> m, v\u1edbi sai s\u1ed1 phép \u0111o tính \u0111\u01b0\u1ee3c là $\\triangle s=0,0031$<\/span> m. \u0110âu là cách vi\u1ebft k\u1ebft qu\u1ea3 \u0111o \u0111úng?<\/p>","options":["A. $s=1,368 \\pm 0,003(m)$<\/span>","B. $s=1,3 \\pm 0,003(m)$<\/span>","C. $s=1,36832 \\pm 0,0031(m)$<\/span>","D. $s=1,3684 \\pm 0,003(m)$<\/span>"],"correct":"1","level":"3","hint":"K\u1ebft qu\u1ea3 \u0111o c\u1ee7a \u0111\u1ea1i l\u01b0\u1ee3ng A cho d\u01b0\u1edbi d\u1ea1ng kho\u1ea3ng giá tr\u1ecb:  $\\bar{A} -\\triangle{A}\\leq A\\leq \\bar{A}+\\triangle{A}$<\/span><\/p>","answer":"Ch\u1ecdn A. Ta có $\\bar{s}=1,36832$<\/span>  m, $\\triangle s=0,0031$<\/span> m K\u1ebft qu\u1ea3 \u0111o \u0111\u01b0\u1ee3c vi\u1ebft: $A=\\bar{A}\\pm \\triangle A$<\/span> , giá tr\u1ecb trung bình $\\bar{A}$<\/span>  \u0111\u01b0\u1ee3c vi\u1ebft \u0111\u1ebfn b\u1eadc th\u1eadp phân t\u01b0\u01a1ng \u1ee9ng v\u1edbi  $ \\triangle A$<\/span> Quy t\u1eafc làm tròn s\u1ed1: n\u1ebfu ch\u1eef s\u1ed1 \u1edf hàng b\u1ecf \u0111i nh\u1ecf h\u01a1n 5 thì ch\u1eef s\u1ed1 bên trái v\u1eabn gi\u1eef nguyên, n\u1ebfu ch\u1eef s\u1ed1 hàng b\u1ecf \u0111i l\u1edbn h\u01a1n 5 thì ch\u1eef s\u1ed1 bên trái t\u0103ng thêm m\u1ed9t \u0111\u01a1n v\u1ecb. => $s=1,368 \\pm 0,003(m)$<\/span>  m<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-09-29 05:30:34","option_type":"math","len":0},{"id":"5658","mon_id":"1158790","chapter_id":"1158791","post_id":"6287","question":"K\u1ebft qu\u1ea3 \u0111o v\u1eadn t\u1ed1c c\u1ee7a m\u1ed9t v\u1eadt r\u01a1i t\u1ef1 do t\u1eeb \u0111\u1ed9 cao h khi ch\u1ea1m \u0111\u1ea5t là $\\nu=10,000 \\pm 0,039 (m\/s)$<\/span> . Hãy tính sai s\u1ed1 t\u1ec9 \u0111\u1ed1i c\u1ee7a phép \u0111o trên.<\/p>","options":["A. $\\delta \\nu= 0,39$<\/span>%","B. $\\delta \\nu= 0,42$<\/span>%","C. $\\delta \\nu= 0,94$<\/span>%","D. $\\delta \\nu= 0,27$<\/span>%"],"correct":"1","level":"3","hint":"Sai s\u1ed1 t\u1ec9 \u0111\u1ed1i c\u1ee7a phép \u0111o là t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m gi\u1eefa sai s\u1ed1 tuy\u1ec7t \u0111\u1ed1i và giá tr\u1ecb trung bình c\u1ee7a \u0111\u1ea1i l\u01b0\u1ee3ng \u0111o.<\/p>","answer":"Ch\u1ecdn A. $\\delta \\nu = \\frac{\\triangle \\nu}{\\nu}.100$<\/span>%$=\\frac{0,039}{10}=0,39$<\/span>%<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2022-09-29 05:43:48","option_type":"math","len":0}]}