Maqolalar

12.3-bo'lim Javoblar - Matematika


1. (u (x, y) = frac {8} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac { sinh (2n-1) pi (1) -y)} {(2n-1) ^ {3} sinh (2n-1) pi} sin (2n-1) pi x )

2. (u (x, y) = - frac {32} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac {(1 + (- 1) ^ {n } 2) sinh n pi (3-y) / 2} {n ^ {3} sinh 3n pi / 2} sin frac {n pi x} {2} )

3. (u (x, y) = frac {8} { pi ^ {2}} sum_ {n = 1} ^ { infty} (- 1) ^ {n + 1} frac { sinh (2n-1) pi (1-y / 2)} {(2n-1) ^ {2} sinh (2n-1) pi} sin frac {(2n-1) pi x} {2} )

4. (u (x, y) = frac { pi} {2} frac { sinh (1-y)} { sinh 1} sin x- frac {16} { pi} sum_ {n = 1} ^ { infty} frac {n sinh 2n (1-y)} {(4n ^ {2} -1) ^ {2} sinh 2n} sin 2nx )

5. (u (x, y) = 3y + frac {108} { pi ^ {3}} sum_ {n = 1} ^ { infty} (- 1) ^ {n} frac { sinh n pi y / 3} {n ^ {3} cosh 2n pi / 3} cos frac {n pi x} {3} )

6. (u (x, y) = frac {y} {2} + frac {4} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac { sinh (2n-1) pi y} {(2n-1) ^ {3} cosh 2 (2n-1) pi} cos (2n-1) pi x )

7. (u (x, y) = - frac {8y} {3} + frac {32} { pi ^ {3}} sum_ {n = 1} ^ { infty} (- 1) ^ {n} frac { sinh n pi y / 2} {n ^ {3} cosh n pi} cos frac {n pi x} {2} )

8. (u (x, y) = frac {y} {3} + frac {4} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac { sinh n pi y} {n ^ {3} cosh n pi} cos n pi x )

9. (u (x, y) = frac {128} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac { cosh (2n-1) pi (x) -3) / 4} {(2n-1) ^ {3} cosh 3 (2n-1) pi / 4} sin frac {(2n-1) pi y} {4} )

10. (u (x, y) = - frac {96} { pi ^ {3}} sum_ {n = 1} ^ { infty} left [1 + (- 1) ^ {n} frac {4} {(2n-1) pi} o'ng] frac { cosh (2n-1) pi (x-2) / 2} {(2n-1) ^ {3} cosh ( 2n-1) pi} sin frac {(2n-1) pi y} {2} )

11. (u (x, y) = frac {768} { pi ^ {3}} sum_ {n = 1} ^ { infty} left [1 + (- 1) ^ {n} frac {2} {(2n-1) pi} right] frac { cosh (2n-1) pi (x-2) / 4} {(2n-1) ^ {3} cosh (2n) -1) pi / 2} sin frac {(2n-1) pi y} {4} )

12. (u (x, y) = frac {96} { pi ^ {3}} sum_ {n = 1} ^ { infty} left [3 + (- 1) ^ {n} frac {4} {(2n-1) pi} o'ng] frac { cosh 3 (2n-1) pi (x-2) / 2} {(2n-1) ^ {3} cosh ( 2n-1) pi / 2} sin frac {(2n-1) pi y} {2} )

13. (u (x, y) = - frac {16} { pi} sum_ {n = 1} ^ { infty} frac { cosh (2n-1) x / 2} {(2n) -3) (2n + 1) (2n-1) sinh (2n-1) / 2} cos frac {(2n-1) y} {2} )

14. (u (x, y) = - frac {432} { pi ^ {3}} sum_ {n = 1} ^ { infty} left [1+ frac {4 (-1) ^ {n}} {(2n-1) pi} right] frac { cosh (2n-1) pi x / 6} {(2n-1) ^ {3} sinh (2n-1) pi / 3} cos frac {(2n-1) pi y} {6} )

15. (u (x, y) = - frac {64} { pi} sum_ {n = 1} ^ { infty} (- 1) ^ {n} frac { cosh (2n-1) ) x / 2} {(2n-1) ^ {4} sinh (2n-1) / 2} cos frac {(2n-1) y} {2} )

16. (u (x, y) = - frac {192} { pi ^ {4}} sum_ {n = 1} ^ { infty} frac { cosh (2n-1) pi x / 2} {(2n-1) ^ {4} sinh (2n-1) pi / 2} left [(- 1) ^ {n} + frac {2} {(2n-1) pi } o'ng] cos frac {(2n-1) pi y} {2} )

17. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} frac { sinh n pi y / a} { sinh n pi b / a} sin frac {n pi x} {a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) sin frac {n pi x} {a} dx, quad u (x, y) = frac {72} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac { sinh (2n- 1) pi y / 3} {(2n-1) ^ {3} sinh 2 (2n-1) pi / 3} sin frac {(2n-1) pi x} {3} )

18. (u (x, y) = alfa_ {0} (1-y / b) + sum_ {n = 1} ^ { infty} alfa_ {n} frac { sinh n pi ( by) / a} { sinh n pi b / a} cos frac {n pi x} {a}, quad alpha_ {0} = frac {1} {a} int_ {0} ^ {a} f (x) dx, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) cos frac {n pi x} { a} dx, quad n geq 1, quad u (x, y) = frac {8 (1-y)} {15} - frac {48} { pi ^ {4}} sum_ { n = 1} ^ { infty} frac { sinh n pi (1-y)} { sinh n pi} cos n pi x )

19. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} frac { sinh (2n-1) pi (by) / 2a} { sinh ( 2n-1) pi b / a} cos frac {(2n-1) pi x} {2a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) cos frac {(2n-1) pi x} {2a} dx, quad u (x, y) = frac {288} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac { sinh (2n-1) pi (2-y) / 6} {(2n-1) ^ {3} sinh (2n-1) pi / 3} sin frac {(2n-1) pi x} {6} )

20. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} frac { sinh (2n-1) pi (by) / 2a} { sinh ( 2n-1) pi b / 2a} sin frac {(2n-1) pi x} {2a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) sin frac {(2n-1) pi x} {2a} dx, quad u (x, y) = frac {32} { pi ^ {3}} sum_ {n = 1} ^ { infty} left [(-1) ^ {n} 5+ frac {18} {(2n-1) pi} right] frac { sinh (2n-1) pi (2-y) / 2} {(2n-1) ^ {3} sinh (2n-1) pi} cos frac {(2n-1) pi x} {2} )

21. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} frac { cosh n pi (yb) / a} { cosh n pi b / a} sin frac {n pi x} {a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) sin frac { n pi x} {a} dx, quad u (x, y) = - 12 sum_ {n = 1} ^ { infty} (- 1) ^ {n} frac { cosh n (y-) 2)} {n ^ {3} cosh 2n} sin nx )

22. (u (x, y) = alfa_ {0} + sum_ {n = 1} ^ { infty} alpha_ {n} frac { cosh n pi y / a} { cosh n pi b / a} cos frac {n pi x} {a}, quad alpha_ {0} = frac {1} {a} int_ {0} ^ {a} f (x) dx , quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) cos frac {n pi x} {a} dx, quad n geq 1, quad u (x, y) = frac { pi ^ {4}} {30} -3 sum_ {n = 1} ^ { infty} frac {1} {n ^ {4}} frac { cosh 2ny} { cos 2n} cos 2nx )

23. (u (x, y) = frac {a} { pi} sum_ {n = 1} ^ { infty} alfa_ {n} frac { sinh n pi (yb) / a } {n cosh n pi b / a} sin frac {n pi x} {a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a } f (x) sin frac {n pi x} {a} dx, quad u (x, y) = frac {4} { pi} sum_ {n = 1} ^ { infty} (-1) ^ {n + 1} frac { sinh (2n-1) (y-1)} {(2n-1) ^ {3} cosh (2n-1)} sin (2n-1) ) x )

24. (u (x, y) = sum_ {n = 1} ^ { infty} alfa _ {n} frac { cosh n pi x / b} { cosh n pi a / b } sin frac {n pi y} {b}, quad alpha_ {n} = frac {2} {b} int_ {0} ^ {b} g (y) sin frac {n pi y} {b} dy, quad u (x, y) = frac {96} { pi ^ {5}} sum_ {n = 1} ^ { infty} frac { cosh (2n -1) pi x} {(2n-1) ^ {5} cosh (2n-1) pi} sin (2n-1) pi y )

25. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} frac { cosh (2n-1) pi x / 2b} { cosh (2n-) 1) pi a / 2b} cos frac {(2n-1) pi y} {2b}, quad alpha_ {n} = frac {2} {b} int_ {0} ^ {b } g (y) cos frac {(2n-1) pi y} {2b} dy, quad u (x, y) = - frac {128} { pi ^ {3}} sum_ { n = 1} ^ { infty} (- 1) ^ {n} frac { cosh (2n-1) pi x / 4} {(2n-1) ^ {3} cosh (2n-1) pi / 2} cos frac {(2n-1) pi y} {4} )

26. (u (x, y) = frac {b} { pi} sum_ {n = 1} ^ { infty} alpha_ {n} frac { cosh n pi x / b} { n sinh n pi a / b} sin frac {n pi y} {b}, quad alpha_ {n} = frac {2} {b} int_ {0} ^ {b} g (y) sin frac {n pi y} {b} dy, quad u (x, y) = frac {64} { pi ^ {3}} sum_ {n = 1} ^ { infty} (- 1) ^ {n + 1} frac { cosh (2n-1) pi x / 4} {(2n-1) ^ {3} sinh (2n-1) pi / 4} sin frac {(2n-1) pi y} {4} )

27. (u (x, y) = - frac {2b} { pi} sum_ {n = 1} ^ { infty} alpha_ {n} frac { cosh (2n-1) pi (xa) / 2b} {(2n-1) sinh (2n-1) pi a / 2b} sin frac {(2n-1) y} {2b}, quad alpha_ {n} = frac {2} {b} int_ {0} ^ {b} g (y) sin frac {(2n-1) pi y} {2b} dy, quad u (x, y) = 192 sum_ {n = 1} ^ { infty} left [1 + (- 1) ^ {n} frac {4} {(2n-1) pi} right] frac { cosh (2n-1) ) (x-1) / 2} {(2n-1) ^ {4} sinh (2n-1) / 2} sin frac {(2n-1) y} {2} )

28. (u (x, y) = alfa_ {0} (xa) + frac {b} { pi} sum_ {n = 1} ^ { infty} alfa_ {n} frac { sinh n pi (xa) / b} {n cosh n pi a / b} cos frac {n pi y} {b}, quad alpha_ {0} = frac {1} {b } int_ {0} ^ {b} g (y) cos frac {n pi y} {b} dy, quad alpha_ {n} = frac {2} {b} int_ {0} ^ {b} g (y) cos frac {n pi y} {b} dy, quad u (x, y) = frac { pi (x-2)} {2} - frac { 4} { pi} sum_ {n = 1} ^ { infty} frac { sinh (2n-1) (x-2)} {(2n-1) ^ {3} cosh 2 (2n-) 1)} cos (2n-1) y )

29. (u (x, y) = alfa_ {0} + sum_ {n = 1} ^ { infty} alfa_ {n} e ^ {- n pi y / a} cos frac { n pi x} {a}, quad alpha_ {0} = frac {1} {a} int_ {0} ^ {a} f (x) dx, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) cos frac {n pi x} {a} dx, quad n geq 1, quad u (x, y) = frac { pi ^ {3}} {2} - frac {48} { pi} sum_ {n = 1} ^ { infty} frac {1} {(2n-1) ^ {4} } e ^ {- (2n-1) y} cos (2n-1) x )

30. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} e ^ {- (2n-1) pi y / 2a} cos frac {(2n) -1) pi x} {2a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) cos frac {(2n-1) pi x} {2a} dx, quad u (x, y) = - frac {288} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac {(- 1 ) ^ {n}} {(2n-1) ^ {3}} e ^ {- (2n-1) pi y / 6} cos frac {(2n-1) pi x} {6} )

31. (u (x, y) = sum_ {n = 1} ^ { infty} alfa_ {n} e ^ {- (2n-1) pi y / 2a} sin frac {(2n) -1) pi x} {2a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) sin frac {(2n-1) pi x} {2a} dx, quad u (x, y) = frac {32} { pi} sum_ {n = 1} ^ { infty} frac {1} {(2n-1) ^ {3}} e ^ {- (2n-1) y / 2} sin frac {(2n-1) x} {2} )

32. (u (x, y) = - frac {a} { pi} sum_ {n = 1} ^ { infty} frac { alpha_ {n}} {n} e ^ {- n pi y / a} sin frac {n pi x} {a}, quad alpha_ {n} = frac {2} {a} int_ {0} ^ {a} f (x) sin frac {n pi x} {a} dx, quad u (x) = 4 sum_ {n = 1} ^ { infty} frac {(1 + (- 1) ^ {n} 2) } {n ^ {4}} e ^ {- ny} sin nx )

33. (u (x, y) = - frac {2a} { pi} sum_ {n = 1} ^ { infty} frac { alpha_ {n}} {2n-1} e ^ { - (2n-1) pi y / 2a} cos frac {(2n-1) pi x} {2a}, quad alpha_ {n} = frac {2} {a} int_ {0 } ^ {a} f (x) cos frac {(2n-1) pi x} {2a} dx, quad u (x, y) = frac {5488} { pi ^ {3}} sum_ {n = 1} ^ { infty} frac {1} {(2n-1) ^ {3}} left [1+ frac {4 (-1) ^ {n}} {(2n-) 1) pi} right] e ^ {- (2n-1) pi y / 14} cos frac {(2n-1) pi x} {14} )

34. (u (x, y) = - frac {2a} { pi} sum_ {n = 1} ^ { infty} frac { alpha _ {n}} {2n-1} e ^ {- (2n-1) pi y / 2a} sin frac {(2n-1) pi x} {2a}, quad alpha_ {n} = frac {2} {a} int_ { 0} ^ {a} f (x) sin frac {(2n-1) pi x} {2a} dx, quad u (x, y) = - frac {2000} { pi ^ {3 }} sum_ {n = 1} ^ { infty} frac {1} {(2n-1) ^ {3}} left [(- 1) ^ {n} + frac {4} {(2n) -1) pi} right] e ^ {- (2n-1) pi y / 10} sin frac {(2n-1) pi x} {10} )

35. (u (x, y) = sum_ {n = 1} ^ { infty}} frac {A_ {n} sinh n pi (by) / a + B_ {n} sinh n pi y / a} { sinh n pi b / a} sin frac {n pi x} {a} + sum_ {n = 1} ^ { infty} frac {C_ {n} sinh n pi (ax) / b + D_ {n} sinh n pi x / b} { sinh n pi a / b} sin frac {n pi y} {b} )

36. (u (x, y) = C + frac {a} { pi} sum_ {n = 1} ^ { infty} frac {B_ {n} cosh n pi y / a-A_ {n} cosh n pi (yb) / a} {n sinh n pi b / a} cos frac {n pi x} {a} + frac {b} { pi} sum_ {n = 1} ^ { infty} frac {D_ {n} cosh n pi x / b-C_ {n} cosh n pi (xa) / b} {n sinh n pi a / b} cos frac {n pi y} {b} )


12.3 stavkalari to'g'risidagi qonunlar

Oldingi modulda aytib o'tilganidek, reaktsiya tezligiga ko'pincha reaktivlarning konsentratsiyasi ta'sir qiladi. Rate qonunlari (ba'zan shunday nomlanadi) differentsial stavka qonunlari) yoki tezlik tenglamalari - bu kimyoviy reaktsiya tezligi va uning reaktivlari kontsentratsiyasi o'rtasidagi bog'liqlikni tavsiflovchi matematik ifodalar. Misol tariqasida kimyoviy tenglama bilan tavsiflangan reaktsiyani ko'rib chiqing

qayerda a va b stexiometrik koeffitsientlardir. Ushbu reaktsiya uchun stavka qonuni quyidagicha yozilgan:

unda [A] va [B] reaktivlarning molyar kontsentratsiyasini ifodalaydi va k ma'lum bir haroratda ma'lum reaktsiya uchun xos bo'lgan tezlik konstantasidir. Eksponentlar m va n reaktsiya tartiblari va odatda musbat butun sonlardir, ammo ular kasrlar, salbiy yoki nol bo'lishi mumkin. Tezlik sobit k va reaktsiya buyruqlari m va n reaktivlarning konsentrasiyalari o'zgarganda reaktsiya tezligi qanday o'zgarishini kuzatish orqali eksperimental tarzda aniqlanishi kerak. Tezlik sobit k reaktiv konsentrasiyalaridan mustaqildir, lekin u haroratga qarab o'zgaradi.

Tezlik qonunidagi reaktsiya tartiblari reaktiv konsentrasiyalariga tezlikning matematik bog'liqligini tavsiflaydi. Yuqoridagi umumiy tezlik qonuniga murojaat qilib, reaktsiya m nisbatan buyurtma A va n nisbatan buyurtma B. Masalan, agar m = 1 va n = 2, reaktsiya birinchi tartibda bo'ladi A va ikkinchi tartib B. Umumiy reaktsiya tartibi shunchaki har bir reaktiv uchun buyurtmalar yig'indisidir. Misol stavkasi qonuni uchun bu erda reaktsiya umumiy tartibda (1 + 2 = 3). Ushbu kontseptsiyani yanada aniqroq ko'rsatish uchun bir nechta aniq misollar quyida keltirilgan.

vodorod peroksidda birinchi darajali va umuman birinchi darajadagi reaktsiyani tavsiflaydi. Stavka qonuni:

C da ikkinchi darajali bo'lgan reaktsiyani tavsiflaydi4H6 va umuman ikkinchi buyurtma. Stavka qonuni:

birinchi darajali H +, birinchi darajali OH - va umuman ikkinchi darajali bo'lgan reaktsiyani tavsiflaydi.

12.3-misol

Reaksiya buyurtmalaridan stavka to'g'risidagi qonunlarni yozish

NO-da ikkinchi tartib2 va 100 ° C da CO da nolinchi tartib. Reaksiya uchun stavka qonuni qanday?

Qaror

Reaksiya NO-da ikkinchi tartibda bo'ladi2 shunday qilib m = 2. Shunday qilib reaktsiya CO da nol tartibda bo'ladi n = 0. Stavka qonuni:

Nolinchi darajaga ko'tarilgan raqam 1 ga teng ekanligini unutmang, shuning uchun [CO] 0 = 1, shuning uchun CO kontsentratsiyasi atamasi tezlik qonunidan chiqarib tashlanishi mumkin: reaktsiya tezligi faqat NO kontsentratsiyasiga bog'liq2. Reaksiya mexanizmlari haqidagi keyingi bobda reaktivning kontsentratsiyasi reaktsiyaga aloqador bo'lishiga qaramay reaksiya tezligiga qanday ta'sir ko'rsatmasligi mumkinligi tushuntiriladi.

O'rganishingizni tekshiring

darajasi = ekanligi aniqlandi k[YO'Q] 2 [H2]. Har bir reaktivga nisbatan qanday tartiblar mavjud va reaktsiyaning umumiy tartibi qanday?

Javob:

NO = 2 tartibida H da tartib2 = 1 umumiy buyurtma = 3

O'rganishingizni tekshiring

Metanol va etil asetat o'rtasidagi reaktsiya tezligi qonuni ma'lum sharoitlarda quyidagicha aniqlanadi:

Metanol va etil asetatga nisbatan reaksiya tartibi qanday va umumiy reaksiya qanday?

Javob:

CHdagi tartib3CHda OH = 1 tartib3CH2OCOCH3 = 0 umumiy buyurtma = 1

Stavka qonunlarini aniqlashda keng tarqalgan eksperimental yondashuv bu boshlang'ich stavkalar usuli. Ushbu usul turli xil boshlang'ich reaktiv kontsentratsiyalari yordamida amalga oshirilgan ko'plab eksperimental sinovlar uchun reaktsiya tezligini o'lchashni o'z ichiga oladi. Ushbu sinovlar uchun o'lchangan stavkalarni taqqoslash reaksiya tartiblarini va keyinchalik stavka qonunini shakllantirish uchun birgalikda ishlatiladigan stavkaning konstantasini aniqlashga imkon beradi. Ushbu yondashuv keyingi ikkita misol mashqlarida tasvirlangan.

12.4-misol

Dastlabki stavkalardan stavka to'g'risidagi qonunni aniqlash

Ushbu reaktsiya laboratoriyada o'rganilgan va 25 ° C da quyidagi tezlik ma'lumotlari aniqlangan.

25 ° C da reaksiya uchun tezlik qonuni va tezlik konstantasini aniqlang.

Qaror

Ning qiymatlarini aniqlang m, nva k quyidagi uch qismli jarayon yordamida eksperimental ma'lumotlardan:

Ning qiymatini aniqlang m [NO] o'zgarib turadigan ma'lumotlardan va [O3] doimiy. So'nggi uchta tajribada [NO] o'zgarib turadi, [O3] doimiy bo'lib qoladi. [NO] 3 dan 4 gacha bo'lgan sinovdan ikki marta ko'payganda, stavka ikki baravar ko'payadi, va 3 dan 5 gacha bo'lgan sinovdan [NO] uch marta ko'payganda, bu ko'rsatkich ham uch baravar ko'payadi. Shunday qilib, stavka [NO] bilan to'g'ridan-to'g'ri proportsionaldir va m stavka qonunida 1 ga teng.

Ning qiymatini aniqlang n ma'lumotlardan [O3] o'zgaradi va [NO] doimiy bo'ladi. Dastlabki uchta tajribada [NO] doimiy va [O3] farq qiladi. Reaksiya tezligi [O ning o'zgarishiga to'g'ridan-to'g'ri mutanosib ravishda o'zgaradi3]. Qachon [O3] 1 dan 2 gacha bo'lgan sinovdan ikki baravar ko'payadi, [O3] 1 dan 3 gacha bo'lgan sinovdan uch baravar ko'payadi, stavka ham uch baravar ko'payadi. Shunday qilib, stavka [O bilan to'g'ridan-to'g'ri proportsionaldir3] va n 1. ga teng, stavka qonuni quyidagicha:

Ning qiymatini aniqlang k kontsentratsiyalarning bir to'plamidan va mos keladigan stavkadan. 1-sinovdan olingan ma'lumotlar quyida keltirilgan:

O'rganishingizni tekshiring

Quyidagi eksperimental ma'lumotlardan reaksiya uchun tezlik qonuni va tezlik konstantasini aniqlang:


12.3 Termodinamikaning ikkinchi qonuni: Entropiya

Ushbu bo'lim oxirida siz quyidagilarni amalga oshirishingiz mumkin:

  • Entropiyani tavsiflang
  • Termodinamikaning ikkinchi qonuniga tavsif bering
  • Termodinamikaning ikkinchi qonuni bilan bog'liq masalalarni echish

O'qituvchilarni qo'llab-quvvatlash

O'qituvchilarni qo'llab-quvvatlash

Ushbu bo'limdagi o'quv maqsadlari o'quvchilaringizga quyidagi standartlarni o'zlashtirishga yordam beradi:

  • (6) Ilmiy tushunchalar. Talaba o'zgarishlar fizik tizim ichida sodir bo'lishini biladi va energiya va impulsning saqlanish qonunlarini qo'llaydi. Talaba:
    • (G) termodinamik qonunlarni, shu jumladan energiyani saqlash va entropiya qonunlarini aks ettiruvchi kundalik misollarni tahlil qilish va tushuntirish.

    Bo'limning asosiy shartlari

    Entropiya

    O'qituvchilarni qo'llab-quvvatlash

    O'qituvchilarni qo'llab-quvvatlash

    [BL] [OL] [AL] Issiqlik va absolyut haroratni ko'rib chiqing. Dvigatel samaradorligi bo'yicha oldingi bahslarni eslang. Talabalarning samaradorlik haqidagi tushunchalarini baholash.

    Bo'limning kirish qismidan eslang, dvigatellarning 100 foiz samaradorligi nazariy jihatdan ham mumkin emas. Ushbu hodisa termodinamikaning ikkinchi qonuni bilan izohlanadi, bu entropiya deb nomlanuvchi tushunchaga asoslanadi. Entropiya - bu tizim buzilishining o'lchovidir. Entropiya shuningdek, energiya qancha ekanligini tasvirlaydi emas ish qilish uchun mavjud. Tizim qanchalik tartibsiz bo'lsa va entropiya qanchalik baland bo'lsa, ish uchun tizimning energiyasi shunchalik kam bo'ladi.

    O'qituvchilarni qo'llab-quvvatlash

    O'qituvchilarni qo'llab-quvvatlash

    Entropiyaning ma'nosini tushunish qiyin, chunki u mavhum tushuncha bo'lib tuyulishi mumkin. Biroq, entropiya misollarini kundalik hayotimizda uchratamiz. Masalan, agar avtomobil shinasi teshilgan bo'lsa, havo har tomonga tarqaladi. Idishdagi suv peshtaxtaga o'rnatilganda, u oxir-oqibat bug'lanib, atrofdagi havoga tarqaladigan alohida molekulalar. Issiq narsa xonaga joylashtirilsa, u tezda issiqlik energiyasini har tomonga yoyadi. Entropiyani energiyaning tarqalishining o'lchovi deb hisoblash mumkin. Bu jarayon davomida qancha energiya tarqalishini o'lchaydi. Har qanday energiya oqimi doimo balanddan pastgacha bo'ladi. Demak, entropiya doimo o'sishga intiladi.

    Ishni bajarish uchun energiyaning barcha shakllaridan foydalanish mumkin bo'lsa-da, mavjud bo'lgan energiyani ish uchun sarflashning iloji yo'q. Binobarin, issiqlik bilan uzatiladigan barcha energiyani ishga aylantirish mumkin emas va ularning bir qismi chiqindi issiqlik shaklida yo'qoladi, ya'ni ish bajarishga ketmaydigan issiqlik. Energiyaning mavjud emasligi termodinamikada muhim ahamiyatga ega, aslida bu maydon dvigatellar tomonidan amalga oshirilgandek, issiqlikni ishlashga aylantirish harakatlaridan kelib chiqqan.

    Rop S Δ S entropiyaning o'zgarishi tenglamasi

    qayerda Q bu jarayon davomida energiyani uzatuvchi issiqlik va T bu jarayon sodir bo'ladigan mutlaq haroratdir.

    Q uzatilgan energiya uchun ijobiy hisoblanadi ichiga tizim issiqlik bilan va energiya uchun salbiy tashqarida tizim issiqlik bilan. SIda entropiya kelvin uchun Jul birliklarida (J / K) ifodalanadi. Agar jarayon davomida harorat o'zgarib tursa, u holda (haroratning ozgina o'zgarishi uchun) odatda yaxshi taxmin qilinadi T Matematikadan (hisob-kitobdan) qochish uchun o'rtacha harorat bo'lishi kerak.

    Muvaffaqiyat uchun maslahatlar

    Mutlaq harorat - bu Kelvinda o'lchangan harorat. Kelvin shkalasi - bu mutlaq noldan yuqori darajalar soni bo'yicha o'lchanadigan mutlaq harorat shkalasi. Shuning uchun barcha harorat ijobiydir. Farengeyt yoki Selsiy kabi boshqa haroratni ishlatib, muttasil bo'lmagan shkala noto'g'ri javob beradi.

    Termodinamikaning ikkinchi qonuni

    Siz hech qachon karta o'yini 52 piketini o'ynaganmisiz? Agar shunday bo'lsa, siz amaliy hazilni boshidan kechirgansiz va bu jarayon davomida siz koinot tabiati to'g'risida termodinamikaning ikkinchi qonuni bilan ta'riflangan qimmatli saboq oldingiz. 52 pikap o'yinida prankster butun kartalarni polga uloqtiradi va siz ularni olib ketishingiz kerak. Kartalarni yig'ish jarayonida siz kartadagi kartani tartibli holatga keltirish uchun zarur bo'lgan ish hajmi kartalarni tashlash va tartibsizlikni yaratish uchun zarur bo'lgan ish hajmidan ancha katta ekanligini sezgan bo'lishingiz mumkin.

    Termodinamikaning ikkinchi qonuni shuni ta'kidlaydi tizimning umumiy entropiyasi har qanday o'z-o'zidan paydo bo'ladigan jarayonda ko'payadi yoki doimiy bo'lib qoladi va u hech qachon kamaymaydi. Ushbu qonunning muhim xulosasi shundaki, issiqlik energiyani o'z-o'zidan yuqori haroratdan past haroratgacha uzatadi, lekin hech qachon o'z-o'zidan teskari yo'nalishda bo'lmaydi. Buning sababi shundaki, energiyani issiqdan sovuqgacha issiqlik uzatishda entropiya kuchayadi (12.9-rasm). Chunki entropiyaning o'zgarishi Q/T, past haroratlarda Δ S Δ S kattaroq o'zgarish bo'ladi (kichikroq) T). Issiq entropiyaning pasayishi (kattaroq) T) ob'ekt sovuqning entropiyasining ko'payishidan kamroq (kichikroq) T) tizim uchun entropiyaning umumiy o'sishini keltirib chiqaradigan ob'ekt.

    Bu haqda o'ylashning yana bir usuli shundaki, har qanday jarayonning yagona natijasi sifatida sovutish moslamasidan issiqroq narsaga issiqlik uzatuvchi energiya bo'lishi mumkin emas. Issiqlik energiyani o'z-o'zidan sovuqdan issiqroq tomon o'tkaza olmaydi, chunki umumiy tizimning entropiyasi pasayadi.

    Birinchidan, nima uchun entropiya ko'paygan? Ikki suv havzasini aralashtirish energiyani yuqori haroratli moddadan past haroratli moddaga issiqlik uzatishi bilan bir xil ta'sirga ega. Aralashtirish issiqroq suvning entropiyasini pasaytiradi, lekin sovuqroq suvning entropiyasini ko'proq ko'paytiradi va entropiyaning umumiy o'sishini keltirib chiqaradi.

    Ikkinchidan, suvning ikki massasi aralashtirilgandan so'ng, issiqlik bilan energiya uzatishni boshqarish va shuning uchun ishni bajarish uchun harorat farqi qolmaydi. Energiya hali ham suvda, ammo hozirda mavjud emas ish qilmoq.

    Uchinchidan, aralash kamroq tartibli yoki boshqa atamani ishlatish uchun kamroq tuzilgan. Har xil haroratda va molekulyar tezliklarning har xil taqsimlanishida ikkita massaga ega bo'lishdan ko'ra, endi bizda molekulyar tezliklarning keng taqsimlanishiga ega bo'lgan bitta massa mavjud, ularning o'rtacha qiymati o'rtacha haroratni beradi.

    Ushbu uchta natija - entropiya, energiyaning yo'qligi va tartibsizlik - nafaqat bog'liqdir, balki aslida tengdir. Energiyani issiqdan sovuqqa issiqlik uzatish tabiatdagi tizimlarning tartibsiz bo'lish tendentsiyasi va ish sifatida foydalanish uchun kamroq energiya mavjudligi bilan bog'liq.

    Ushbu qonunga asoslanib, nima bo'lishi mumkin emas? Issiq bilan aloqa qiladigan sovuq narsa hech qachon o'z-o'zidan energiyani issiqlik bilan issiq narsaga o'tkazmaydi va issiq narsa qizib ketganda soviydi. Shuningdek, issiq, harakatsiz avtomobil hech qachon o'z-o'zidan sovib ketmaydi va harakatlana olmaydi.

    Yana bir misol - vakuum kamerasining bir burchagiga kiritilgan gaz pufagining kengayishi. Gaz kamerani to'ldirish uchun kengayadi, lekin u hech qachon burchakda o'z-o'zidan qayta to'planmaydi. Gaz molekulalarining tasodifiy harakati ularning hammasini burchakka qaytarishi mumkin edi, ammo bu hech qachon kuzatilmaydi (12.10-rasm).

    Biz issiqlik hech qachon energiyani o'z-o'zidan sovuqdan issiqroq narsaga o'tkazmasligini tushuntirdik. Bu erda asosiy so'z o'z-o'zidan. Agar biz ish qil tizimda, u bu energiyani issiqlik bilan sovuqroqdan issiqroq narsaga o'tkazish mumkin. Bu haqda ko'proq ma'lumotni keyingi bobda bilib olamiz, bu termodinamika qonunlarining qo'llanilishlaridan biri sifatida muzlatgichlarni o'z ichiga oladi.

    Ba'zida odamlar termodinamikaning ikkinchi qonunini noto'g'ri tushunishadi, chunki ushbu qonunga asoslanib, entropiyaning biron bir joyda kamayishi mumkin emas. Ammo, aslida bu entropiyasi uchun mumkin bitta qism koinotning kamayishi, olam entropiyasining umumiy o'zgarishi kuchaygan ekan. Tenglama shaklida biz buni quyidagicha yozishimiz mumkin

    Δ S tot = Δ S syst + Δ S envir & gt 0. Δ S tot = Δ S syst + Δ S envir & gt 0.

    Tizim entropiyasi qanday kamayishi mumkin? Energiya uzatish zarur. Agar siz xonada tarqalgan marmarlarni olib, stakanga solib qo'ysangiz, sizning ishingiz ushbu tizim entropiyasini kamaytirgan. Agar siz temir rudasini yerdan yig'ib, uni po'latga aylantirsangiz va ko'prik qursangiz, sizning ishingiz ushbu tizimning entropiyasini kamaytirdi. Quyoshdan keladigan energiya Yerdagi mahalliy tizimlarning entropiyasini kamaytirishi mumkin, ya'ni Δ S syst Δ S sistasi manfiydir. Ammo qolgan koinotning umumiy entropiyasi ko'proq kattalashadi - ya'ni Δ S atrof Δ S atrof ijobiy va kattaligi jihatidan katta. Temir javhari masalasida, garchi siz ko'prik va po'lat tizimini yanada tizimli qilgan bo'lsangiz ham, siz buni olam hisobiga qildingiz. Umuman olganda, olamning entropiyasi rudani qazib olish va uni po'latga aylantirish natijasida yuzaga keladigan tartibsizlik bilan kuchayadi. Shuning uchun,

    va termodinamikaning ikkinchi qonuni emas buzilgan.

    Har safar o'simlik quyosh energiyasini kimyoviy potentsial energiya shaklida saqlaganida yoki iliq havoning yangilanishi parvoz qilayotgan qushni ko'targanida, Yer Quyoshdan chuqur kosmosga energiya uzatishning bir qismini ishlatib, ishlarni bajarish uchun entropiyaning kamayishini sezadi. . Ushbu katta energiya uzatilishi natijasida entropiyaning umumiy o'sishi kuzatiladi. Ushbu energiya uzatilishining kichik bir qismi Yerdagi tuzilgan tizimlarda saqlanadi, natijada entropiyaning ancha kichik, mahalliy pasayishiga olib keladi.

    O'qituvchilarni qo'llab-quvvatlash

    O'qituvchilarni qo'llab-quvvatlash

    [AL] Talabalardan termodinamikaning ikkinchi qonuni to'g'ri bo'lmagan taqdirda nima bo'lishini so'rang. Agar energiya oqimining yo'nalishini oldindan aytib bo'lmaydi bo'lsa-chi? Er yuzidagi hayot o'z vazifasini bajara oladimi?

    Termodinamikaning ikkinchi qonuni bilan bog'liq muammolarni hal qilish

    Entropiya nafaqat energiyani ishlashga yaroqsizligi bilan bog'liq, balki tartibsizlik o'lchovidir. Masalan, muzning erishi blokida yuqori darajada tuzilgan va tartibli suv molekulalari tizimi tartibsiz suyuqlikka aylanadi, unda molekulalar qat'iy pozitsiyalarga ega emas (12.11-rasm). Ushbu jarayon uchun entropiyaning ko'payishi kuzatilmoqda, chunki biz quyidagi ish misolida ko'rib chiqamiz.

    Ishlagan misol

    Buzuqlik bilan bog'liq entropiya

    Dastlab 0 ° C 0 ° C da bo'lgan va 0 ° C 0 ° C da suv hosil qiladigan eriydigan 1,00 kg muzning entropiyasining ko'payishini toping.

    Strategiya

    Entropiyaning o'zgarishini energiyani topgandan so'ng Δ S Δ S ta'rifidan hisoblash mumkin, Q, muzni eritishi kerak edi.

    Entropiyaning o'zgarishi quyidagicha aniqlanadi

    Bu yerda, Q 1,00 kg muzni eritish uchun zarur bo'lgan issiqlik va u tomonidan beriladi

    Chunki Q bu muzga qo'shadigan energiya issiqlik miqdori, uning qiymati ijobiy va T muzning erishi harorati, T = 273 K T = 273 K Demak, entropiyaning o'zgarishi

    Entropiyaning o'zgarishi ijobiy, chunki issiqlik energiyani uzatadi ichiga o'zgarishlar o'zgarishiga olib keladigan muz. Bu entropiyaning sezilarli darajada ko'payishi, chunki u nisbatan past haroratda sodir bo'ladi. Bu suv molekulalarining buzilishining ko'payishi bilan birga keladi.

    Amaliy muammolar

    1. 1.84 marta 10 ^ <3> , matn
    2. 3.67 marta 10 ^ <3> , matn
    3. 1.84 marta 10 ^ <8> , matn
    4. 3.67 marta 10 ^ <8> , matn

    Tushunishingizni tekshiring

    O'qituvchilarni qo'llab-quvvatlash

    O'qituvchilarni qo'llab-quvvatlash

    Ushbu savollardan talabalarning bo'limning o'quv maqsadlariga erishishini baholash uchun foydalaning. Agar talabalar aniq maqsad bilan kurashayotgan bo'lsa, bu savollar qaysi va qaysi talabalarni tegishli tarkibga yo'naltirishini aniqlashga yordam beradi.

    1. Entropiya - bu tizimning potentsial energiyasining o'lchovidir.
    2. Entropiya - bu tizim tomonidan bajarilgan aniq ishning o'lchovidir.
    3. Entropiya - bu tizim buzilishining o'lchovidir.
    4. Entropiya - energiyaning tizimga issiqlik uzatish o'lchovidir.

    Ishni bajarish uchun energiyaning qaysi shakllaridan foydalanish mumkin?

    1. Faqat ishgina ishni bajarishga qodir.
    2. Faqat issiqlik ish qilishga qodir.
    3. Faqat ichki energiya ish qilishga qodir.
    4. Issiqlik, ish va ichki energiya barchasi ishni bajarishga qodir.
    1. Barcha o'z-o'zidan paydo bo'ladigan jarayonlar tizimning umumiy entropiyasining pasayishiga olib keladi.
    2. Barcha o'z-o'zidan paydo bo'lgan jarayonlar tizimning umumiy entropiyasini kuchayishiga olib keladi.
    3. Barcha o'z-o'zidan paydo bo'ladigan jarayonlar tizimning pasayishi yoki doimiy ravishda umumiy entropiyasini keltirib chiqaradi.
    4. Barcha o'z-o'zidan paydo bo'ladigan jarayonlar tizimning doimiy entropiyasini kuchayishiga yoki doimiy ravishda rivojlanishiga olib keladi.

    Energiyani yuqori haroratdan past haroratgacha uzatish uchun odatda butun tizimning entropiyasi nima bo'ladi?

    1. U kamayadi.
    2. U doimiy bo'lib qolishi kerak.
    3. Tizimning entropiyasini harorat uchun aniq qiymatlarsiz oldindan aytib bo'lmaydi.
    4. U ko'payadi.

    Amazon Associate sifatida biz malakali xaridlardan daromad olamiz.

    Ushbu kitobni keltirib, baham ko'rishni yoki o'zgartirmoqchimisiz? Ushbu kitob Creative Commons Attribution License 4.0 bo'lib, siz Texas Education Agency (TEA) ga tegishli bo'lishingiz kerak. Materialning asl nusxasi: https://www.texasgateway.org/book/tea-physics. Asl materialga o'zgartirishlar kiritildi, shu jumladan badiiy, tuzilish va boshqa tarkibni yangilash.

      Agar siz ushbu kitobni to'liq yoki bir qismini bosma shaklda qayta tarqatayotgan bo'lsangiz, unda har bir jismoniy sahifaga quyidagi atributni kiritishingiz kerak:

    • Iqtibos yaratish uchun quyidagi ma'lumotlardan foydalaning. Bu kabi iqtibos vositasidan foydalanishni tavsiya etamiz.
      • Mualliflar: Pol Piter Urone, Rojer Xinrixs
      • Nashriyotchi / veb-sayt: OpenStax
      • Kitob nomi: Fizika
      • Nashr qilingan sana: 2020 yil 26 mart
      • Manzil: Xyuston, Texas
      • Kitobning URL manzili: https://openstax.org/books/physics/pages/1-introduction
      • Bo'limning URL manzili: https://openstax.org/books/physics/pages/12-3-second-law-of-thermodynamics-entropy

      © Yanvar 29, 2021 Texas Ta'lim Agentligi (TEA). OpenStax nomi, OpenStax logotipi, OpenStax kitob muqovalari, OpenStax CNX nomi va OpenStax CNX logotipi Creative Commons litsenziyasiga kirmaydi va Rays Universitetining oldindan va aniq yozma roziligisiz ko'paytirish mumkin emas.


      Gemdas ishchi varag'i - Qaror

      Quyidagi ifodani baholang

      & # xa0 = & # xa0 & # xa0 5 & ​​# xa0 5 & # xa0 - & # xa016 & # xa0 (Ko'paytirishni bajaring, 5 (5) & # xa0 = 25)

      & # xa0 = & # xa0 & # xa0 25 - & # xa0 16 (Ayirishni bajaring)

      Quyidagi ifodani baholang

      = & # xa0 2 [5 & # xa0 + & # xa0(30 & # xa0 ÷ 6) ²] (ichki qavsni baholang)

      = & # xa0 2 [5 & # xa0 + & # xa05²] (Quvvatni baholang)

      Quyidagi ifodani baholang

      = & # xa0 & # xa0 (6 + 4 ²) / (3 ² & # xa0 ⋅ 4) (kuchlarni baholang)

      Quyidagi ifodani baholang

      Quyidagi ifodani baholang

      Ichki qavsdagi belgilarni baholash,

      Chapdan o'ngga, bizda birinchi navbatda ko'paytma mavjud. Shunday qilib, biz 4 ni 17 ga ko'paytirib, keyin 2 ga bo'lishimiz kerak.

      Ikkala raqamni va maxrajni ikkiga bo'lish, biz olamiz

      Yuqorida keltirilgan narsalarni ko'rib chiqib, talabalar "Gemdas ishchi varag'i javoblari bilan" tushungan bo'lishlariga umid qilamiz. & # xa0

      Yuqorida keltirilgan narsalardan tashqari, "Gemdas ishchi varag'i javoblari" haqida ko'proq bilmoqchi bo'lsangiz, iltimos, shu yerni bosing

      Ushbu bo'limda keltirilgan narsalardan tashqari, agar sizga matematikadan boshqa narsalar kerak bo'lsa, iltimos, bizning Google maxsus qidiruvimizdan foydalaning.

      Agar bizning matematik tarkibimiz haqida fikringiz bo'lsa, iltimos, bizga elektron pochta orqali xabar yuboring: & # xa0

      Fikr-mulohazangizni har doim minnatdormiz. & # Xa0

      Siz matematikadan turli xil narsalar bo'yicha quyidagi veb-sahifalarga tashrif buyurishingiz mumkin. & # Xa0


      Oregon shtatidagi matematik irqchilik g'oyasini ilgari surdi va o'qituvchilarni oq ustunlikni yo'q qilishga undaydi

      O'tgan hafta kafedra "Matematikada o'qitishdagi irqchilikni demontaj qilish" bo'limini o'z ichiga olgan & # 8220A Teng matematik o'qitish yo'li "deb nomlangan kurs uchun onlayn risola yubordi. Risolaga asosan xonalar ichida uyg'oqlikni tarqatish uchun mo'ljallangan materiallar kiritilgan.

      ODE tomonidan targ'ib qilingan kurs o'zini "6-8 sinf o'quvchilari qora, lotin va ko'p tilli o'quvchilarni markazlashtiradigan, matematik tenglik yo'lidagi to'siqlarni hal qiladigan va darslarni ustuvor standartlarga moslashtiradigan matematikaga yaxlit yondashuv" deb nomlaydi. Kurs haqiqatan ham ba'zi javoblarning shunchaki noto'g'riligini va haqiqatan ham Amerikani tashkil etish tamoyillarini qiyinlashtirmoqda.

      Kurs dasturiga ko'ra, o'qituvchilarga o'z sinflarida matematikadan chetlashtirishga yordam beradigan beshta & # 8220stride & # 8221 mavjud. Parchalanayotgan irqchilik bo'limida kurs o'qituvchilardan ularning "matematikani o'qitish atrofidagi harakatlari, e'tiqodlari va qadriyatlarini" tekshirishni so'raydi, chunki ularni identifikatsiya qilish siyosati to'g'risida darslar berish uchun ularni sharmanda qilish.

      "Oq ustunlik madaniyati o'qituvchilarning kundalik harakatlarida matematika xonalariga kirib boradi. Ushbu harakatlar asosida yotgan e'tiqodlar bilan bir qatorda, ular [ozchilik] o'quvchilariga matematik olamga to'liq kirish imkoniyatidan mahrum bo'lib, ta'limga zarar etkazadilar.

      Kurs qismida, shuningdek, "matematika sinflarida oq ustunlik paydo bo'lishi usullari" aniqlangan.

      O'quv materiallariga ko'ra, matematik masalalarni echishda talabalardan "to'g'ri" javob olishni so'rashga juda katta e'tibor, mustaqil ravishda ishlash kabi oq ustunlikka bog'liqdir. Kurs talabalar uchun matematikadan foydalanishni ma'qullaydi.

      "Ko'pincha" haqiqiy dunyoda "matematikani o'rganishga ahamiyat beriladi, go'yo bizning sinflarimiz haqiqiy dunyoning bir qismi emas. Bu yoki / yoki fikrlash tushunchalarini kuchaytiradi, chunki matematika faqat ma'lum bir kontekstda bo'lganida foydali bo'ladi », - deyiladi demontaj qilinayotgan irqchilik qismida. "Biroq, bu dunyodagi mavjudot va tushunchalarni kapitalistik va imperialistik usullarini qo'llab-quvvatlash uchun matematikadan foydalanishga olib kelishi mumkin".

      Kurs "kapitalizm" va "imperializm" ga qaratilgan bo'lib, matematik darslarni kam ta'minlangan bolalar mahallalarini rivojlantirishga yordam berish uchun taklif qiladi.

      Materiallar, shuningdek, o'qituvchilarni "matematikaning o'quvchilar jamoalariga ta'sir qiladigan muammolarni hal qilishda qanday yordam berishini" so'rashga va "rang-barang jamoalarning matematikaga jalb qilish va kundalik hayotida muammolarni hal qilish usullarini" tan olishga taklif qiladi.

      Bundan tashqari, kurs o'qituvchilardan "matematikadan kapitalistik, imperialistik va irqchi qarashlarni qo'llab-quvvatlashda foydalaniladigan usullarni" aniqlash va ularga qarshi chiqishlarini so'raydi.

      & # 8220Matematika kontseptsiyasi mutlaqo ob'ektiv ekanligi shubhasiz yolg'ondir va uni o'rgatish bundan ham kamroq, & # 8221 kursda ta'kidlangan. & # 8220Hamma to'g'ri va noto'g'ri javoblar bor degan fikrni qo'llab-quvvatlash ob'ektivlikni kuchaytiradi, shuningdek ochiq mojarolardan qo'rqadi. "

      Kurs marksistik ideallarni targ'ib qilishning eng yaxshi usullarini o'rganishni istagan o'qituvchilar uchun ixtiyoriy bo'lib ko'rinadi. Ammo bu steroidlar bo'yicha tanqidiy poyga nazariyasi² & # 8211 1.776 + 1.619.

      Masalan, marksistik materiallar "Biz odamlarni xususiy foyda olish uchun ishlatadigan tizimda irqchilikni tarqatib yuborolmaymiz & # 8230 Agar biz irqchilikni yo'q qilmoqchi bo'lsak, unda iqtisodiy adolat uchun harakatni qurishimiz kerak" deb ta'kidlaydi.

      Bu mamlakat ta'lim tizimining davlati yoki hech bo'lmaganda kelajagi. Mamlakatimizdagi shaharlardagi o'qituvchilar hozirgi paytda maktablarni yopib qo'yish uchun koronavirus pandemiyasi yordamida tumanlarni garovga olishmoqda. Once many of those teachers finally get around to doing their jobs, some kids will likely be further underserved by lessons that are not relevant to promoting critical thinking — at least in the Beaver State.

      The course, more or less, seems to be an attack on the fundamentals of the country while using objective number-crunching as a pretext. It also infers that minorities do not have the ability to solve problems — which itself seems kind of racist.

      In Oregon, the idea that there are wrong answers to math problems is being challenged, as is the notion that minority kids are capable of competing in STEM activities without assistance.

      Apparently every lesson will be one where slackers all sit around the smart kid waiting for that person to come up with the correct answer. Weren’t those rare group opportunities fun?

      It isn’t clear, beyond the Marxist-rhetoric-serving adults, how students are supposed to actually benefit from the absurd social experiment that the course is.

      If you’re wondering where something so brazenly anti-American originated from, the lesson thanks the Bill and Melinda Gates Foundation for its “generous financial support.”


      Detailed Answer Key

      Find the simple interest on $6,900 at 16 ⅔% per year ਏor 2 years.

      Oddiy qiziqish uchun formulalar

      Plug these values in the above formula

      Hence, the interest earned is $240.

      If a sum of money is doubled in 10 years in simple interest, in how many years will it be tripled ? 

      Let P be the sum of money. 

      Berilgan:  P is doubled in 10 years

      Endi biz foizlarni o'n yil davomida quyida keltirilgan tarzda hisoblashimiz mumkin & # xa0

      Yuqoridagi hisob-kitoblarga ko'ra, P - bu dastlabki 10 yil uchun foizlar. & # Xa0

      Oddiy foizlarda, olingan foiz har yili uchun bir xil bo'ladi.

      Interest earned in the next 10 years will also be P. 

      Bu quyida tushuntirilgan.

      Hence, the sum of money will be tripled in 20 years. 

      If a sum of money amounts to $ 6200 in 2 years and $ 7400 in 3 years under simple interest, then find the principal.

      From the given given information we have the following points.

      2 yil oxirida biz $ 6200 olamiz

      3 yil oxirida biz 7400 dollar olamiz

      From the above two points,  we can get the interest earned in the 3rd year. 

      It has been explained below. 

      Oddiy qiziqish uchun foiz har yili uchun bir xil bo'ladi.

      So, we can calculate the principal as given below.

      Demak, asosiy qarz $ 3800.

      If a sum of money produces $3900 as interest in 3 years and 3 months at 16% per year simple interest, find the principal.  

      Oddiy qiziqish uchun formulalar

      The value of "t" must always be in "years". But in the question, it is given in both years and months. 

      To convert months into years, we have to divide the given months by 12.

      Multiply both sides by 25/13.

      Hence, the required principal is $7500. 

      Arthur invests his inheritance of $24,000 in two different accounts which pay 6% and 5% annual interest. After one year, he received $1330 in interest. How much did he invest in each account ?  

      Let "x" be the amount invested in 6% account. 

      Then, the amount invested in 5% account is

      Berilgan:  Total interest earned on both the accounts is $1340.

      Interest in 6% account + Interest in 5% account  =  1330

      Subtract 1200 from both sides

      Hence, the amount invested in 6% account is $13,000 and in 5% account is $11,000.

      Mr. Garret invested twice as much money at 6% as he did at 7%. After one year, his earnings at 6% were $95 more than his earnings at 7%. Find the amount invested at each rate.   

      Let "x" be the amount invested at 7% rate. 

      Then, the amount invested in 6% rate is

      Interest earned after 1 year at 7% rate is 

      Interest earned after 1 year at 6% rate is 

      Berilgan: Earnings at 6% were $95 more than his earnings at 7%. 

      That is, earnings in (2) were $95 moire than the earnings in (1). 

      Hence, the amount invested at 7% rate $1900 and at 6% rate is $3800. 

      Apart from the stuff given in this section , if  you need any other stuff in math, please use our google custom search here.

      Agar bizning matematik tarkibimiz haqida fikringiz bo'lsa, iltimos, bizga elektron pochta orqali xabar yuboring: & # xa0

      Fikr-mulohazangizni har doim minnatdormiz. & # Xa0

      Siz matematikadan turli xil narsalar bo'yicha quyidagi veb-sahifalarga tashrif buyurishingiz mumkin. & # Xa0


      Section 12.3 Answers - Mathematics

      Inscribed angles

      Before we begin, let’s state a few important theorems.

      THEOREM: If two angles inscribed in a circle intercept the same arc, then they are equal to each other.

      THEOREM: If an angle inside a circle intercepts a diameter, then the angle has a measure of (90^circ ).

      Now let’s use these theorems to find the values of some angles!

      EXAMPLE: Find the measure of the angle indicated.

      SOLUTION: First we see that (measuredangle DNC) and (measuredangle DEC) both intercept the same arc. So, by our first theorem, we know that (measuredangle DNC = measuredangle DEC). And since (measuredangle DNC = 37^circ ), we conclude that (measuredangle DEC = 37^circ ).

      Next, since (measuredangle EDC) intercepts a diameter of the circle, we can conclude, by our second theorem, that (measuredangle EDC = 90^circ ).

      So two of the angles in triangle (EDC) are (90^circ ) and (37^circ ). Then, by using the fact that the tree angles of a triangle must sum to (180^circ ), we know that the measure of the desired angle is (53^circ ).

      EXAMPLE: Find the measure of the indicated angle.

      SOLUTION: Since (measuredangle NBM) and (measuredangle NLM) intercept the same arc, they are equal, so that (measuredangle NLM = 52^circ ). Next, since (measuredangle NML) intersects a diameter of the circle, we know that (measuredangle NML = 90^circ ). Then since two of the angles of triangle (LMN) are (90^circ ) and (52^circ ), we know that the desired angle must be (38^circ ), by the fact that the three angles of a triangle must sum to (180^circ ). Solution: The desired angle is (38^circ ).

      Below you can yuklab olish biroz ozod math worksheets and practice.


      This question requires you to write out all the steps, even though the math itself isn't too complicated.

      You're trying to figure out the price per pound of beef (b) when it was equal to the price per pound of chicken (c). In other words, when b = c, or 2.35 + 0.25x = 1.75 + 0.40x. So you need to find the value of x in order to plug it back into the "b" equation, writes Dora Seigel of PrepScholar.

      Subtract 1.75 from each side:

      2.35(−1.75) + 0.25x = 1.75(−1.75) + 0.40x

      That leaves you with 0.6 + 0.25x = 0.40x. So subtract 0.25x from each side:

      The last step is to reduce the equation:

      Now that you know the value of x, you can put it into the equation for the price of beef:

      The correct answer is "D," $3.35.


      General Knowledge

      The Mathematics Subtest consists of approximately 40 multiple-choice questions and is 100 minutes long.

      Each of the questions will contain four response options. You will choose the best response out of the available options, and indicate A, B, C, yoki D.. For the Mathematics Subtest, an on-screen 4-function calculator and a mathematics reference sheet will be provided.

      The table below presents types of questions on the exam and directs you to examples of these formats among the sample items that follow.

      Table of Question Formats

      Type of Question Sample Question
      Scenario
      Examine a situation, problem, or case study. Then answer a question, make a diagnosis, or recommend a course of action by selecting the best response option.
      Question 2
      Word Problem
      Apply mathematical principles to solve a real-world problem.
      Question 7
      Direct Question
      Choose the response option that best answers the question.
      Question 18
      Command
      Select the best response option.
      Question 19
      Graphics
      Choose the option that best answers a question involving a number line, a geometric figure, graphs of lines or curves, a table, or a chart.
      Question 34

      Savollarning namunalari

      The following questions represent both the form and content of questions on the examination. These questions will acquaint you with the general format of the examination however, these sample questions do not cover all of the competencies and skills that are tested and will only approximate the degree of examination difficulty.


      Geometry: Answer Key

      This provides the answers and solutions for the Put Me in, Coach! exercise boxes, organized by sections.

      Taking the Burden out of Proofs

      1. Ha
      2. Theorem 8.3: If two angles are complementary to the same angle, then these two angles are congruent.

      ?A and ?B are complementary, and ?C and ?B are complementary.

      Given: ?A and ?B are complementary, and ?C and ?B are complementary.

      Proving Segment and Angle Relationships

      Given: E is between D and F

      StatementsReasons
      1.E is between D and FBerilgan
      2.D, E, and F are collinear points, and E is on DFDefinition of between
      3.DE + EF = DFSegment Addition Postulate
      4.DE = DF ? EFSubtraction property of equality

      2. If ?BD divides ?ABC into two angles, ?ABD and ?DBC, then m?ABC = m?ABC - m?DBC.

      ?BD divides ?ABC into two angles, ?ABD and ?DBC.

      Given: ?BD divides ?ABC into two angles, ?ABD and ?DBC

      StatementsReasons
      1.?BD divides ?ABC into two angles, ?ABD and ?DBCBerilgan
      2.m?ABD + m?DBC = m?ABCAngle Addition Postulate
      3.m?ABD = m?ABC - m?DBCSubtraction property of equality

      3. The angle bisector of an angle is unique.

      ?ABC with two angle bisectors: ?BD and ?BE.

      Given: ?ABC with two angle bisectors: ?BD and ?BE.

      4. The supplement of a right angle is a right angle.

      ?A and ?B are supplementary angles, and ?A is a right angle.

      Given: ?A and ?B are supplementary angles, and ?A is a right angle.

      StatementsReasons
      1.?A and ?B are supplementary angles, and ?A is a right angleBerilgan
      2.m?A + m?B = 180Definition of supplementary angles
      3.m?A = 90Definition of right angle
      4.90 + m?B = 180Substitution (steps 2 and 3)
      5.m?B = 90Algebra
      6.?B is a right angleDefinition of right angle

      Proving Relationships Between Lines

      1. m?6 = 105 , m?8 = 75
      2. Theorem 10.3: If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent.

      l ? ? m cut by a transversal t.

      Given: l ? ? m cut by a transversal t.

      3. Theorem 10.5: If two parallel lines are cut by a transversal, then the exterior angles on the same side of the transversal are supplementary angles.

      l ? ? m cut by a transversal t.

      Given: l ? ? m cut by a transversal t.

      Prove: ?1 and ?3 are supplementary.

      Interested in 3D Printing?

      We?ve researched the top things you should consider when purchasing a 3Dprinter, and have chosen the best printers of 2020 based on your needs.

      4. Theorem 10.9: If two lines are cut by a transversal so that the alternate exterior angles are congruent, then these lines are parallel.

      Lines l and m are cut by a transversal t.

      Given: Lines l and m are cut by a transversal t, with ?1

      5. Theorem 10.11: If two lines are cut by a transversal so that the exterior angles on the same side of the transversal are supplementary, then these lines are parallel.

      Lines l and m are cut by a t transversal t.

      Given: Lines l and m are cut by a transversal t, ?1 and ?3 are supplementary angles.

      Two's Company. Three's a Triangle

      Given: ?ABC is a right triangle, and ?B is a right angle.

      Prove: ?A and ?C are complementary angles.

      StatementReasons
      1.?ABC is a right triangle, and ?B is a right angleBerilgan
      2.m?B = 90Definition of right angle
      3.m?A + m?B + m?C = 180Theorem 11.1
      4.m?A + 90 + m?C = 180Substitution (steps 2 and 3)
      5.m?A + m?C = 90Algebra
      6.?A and ?C are complementary anglesDefinition of complementary angles

      3. Theorem 11.3: The measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles.

      ?ABC with exterior angle ?BCD.

      StatementReasons
      1.?ABC with exterior angle ?BCDBerilgan
      2.?DCA is a straight angle, and m?DCA = 180Definition of straight angle
      3.m?BCA + m?BCD = m?DCAAngle Addition Postulate
      4.m?BCA + m?BCD = 180Substitution (steps 2 and 3)
      5.m?BAC + m?ABC + m?BCA = 180Theorem 11.1
      6.m?BAC + m?ABC + m?BCA = m?BCA + m?BCDSubstitution (steps 4 and 5)
      7.m?BAC + m?ABC = m?BCDSubtraction property of equality

      6. No, a triangle with these side lengths would violate the triangle inequality.

      Congruent Triangles

      Symmetric property: If ?ABC

      Transitive property: If ?ABC

      = ?DCB as shown in Figure 12.5, then ?ACB

      = ?DCB, as shown in Figure 12.8, then ?ACB

      = ?CDB, as shown in Figure 12.10, then ?ACB

      = CD, as shown in Figure 12.12, then ?ACB

      = ?R and M is the midpoint of PR, as shown in Figure 12.17, then ?N

      Smiliar Triangles

      Opening Doors with Similar Triangles

      1. If a line is parallel to one side of a triangle and passes through the midpoint of a second side, then it will pass through the midpoint of the third side.

      DE ? ? AC and D is the midpoint of AB.

      Given: DE ? ? AC and D is the midpoint of AB.

      Prove: E is the midpoint of BC.

      2. AC = 4?3 , AB = 8? , RS = 16, RT = 8?3

      Putting Quadrilaterals in the Forefront

      Trapezoid ABCD with its XB CY four altitudes shown.

      3. Theorem 15.5: In a kite, one pair of opposite angles is congruent.

      4. Theorem 15.6: The diagonals of a kite are perpendicular, and the diagonal opposite the congruent angles bisects the other diagonal.


      Videoni tomosha qiling: Предел последовательности #25 limann!=0 (Dekabr 2021).