Физика

Mavzu №3. Masalalar yechish. Mavzular kesimida amaliyotga yo’naltirilgan va fanlararo bog’liqlikga doir masalalar yechish

Reja:
Harakatlarning mustaqillik prinsipi
Jismning vertikal harakati
Tekis o’zgaruvchan aylanma harakatda burchak tezlik va tezlanish, normal va tangensal tezlanish.

Harakatlarning mustaqillik prinsipi

Jismning vertikal harakati

Aylanma harakat

Maxovik aylanganda gardishidagi nuqtalar tezligi
6 m/s ulardan o‘qqa 1,5 sm yaqin masofada bo‘lgan nuqtalar tezligi esa 5,5 m/s bo‘lsa, maxovikning radiusi qancha? (Javobi: 1,63 sm).

r

R

Motorli qayiq daryoda manzilga yetib borish uchun 1,8 soat, qaytib kelish uchun esa 2,4 soat vaqt sarfladi. Agar sol jo‘natilsa, manzilga qancha vaqtda yetib boradi? (Javobi: 14,4 soat).
Metrodagi eskalator odamni 30 s da yuqoriga olib chiqadi. Agar odam va eskalator birgalikda harakat qilsa, 10 s da ko‘tariladi. Eskalator tinch tursa odam qancha vaqtda yuqoriga chiqadi? (Javobi: 15 s).
Jism 80 m balandlikdan erkin tushmoqda. Tushishning oxirgi sekundidagi ko‘chishni toping. Harakat davomidagi o‘rtacha tezligini aniqlang. Jismning boshlang‘ich tezligini nolga teng deb hisoblang. (Javobi: 35 m, 20 m/s).
Agar vertikal yuqoriga otilgan jism yo‘lning oxirgi 1/4 qismini 3 s da bosib o‘tgan bo‘lsa, u qancha vaqt ko‘tarilgan? Uning boshlang‘ich tezligi qanday bo‘lgan? (Javobi: 60 m/s, 6 s).
Agar boshlang‘ich tezliksiz erkin tushayotgan jism oxirgi sekundda
75 m yo‘lni o‘tgan bo‘lsa, u qanday balandlikdan tushgan? Harakatning oxiridagi tezligi nimaga teng? (Javobi: 320 m, 80 m/s).
Ikki sharcha bir nuqtadan 20 m/s boshlang‘ich tezlik bilan 1 sekund vaqt intervali bilan yuqoriga vertikal otildi. Birinchi sharcha otilgandan qancha vaqt o‘tgach, sharlar uchrashadi? (Javobi: 2,5 s).

Na’muna

1. Motorli qayiq daryoda manzilga yetib borish uchun 1,8 soat, qaytib kelish uchun esa 2,4 soat vaqt sarfladi. Agar sol jo‘natilsa, manzilga qancha vaqtda yetib boradi? (Javobi: 14,4 soat).

𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 𝜗 𝑜 𝜗𝜗 𝜗 𝑜 𝑜𝑜 𝜗 𝑜 + 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑞 − 𝜗 𝑜 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 − 𝜗 𝑜 𝜗𝜗 𝜗 𝑜 𝑜𝑜 𝜗 𝑜 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8=𝑆 𝜗 𝑞 − 𝜗 𝑜 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8= 𝜗 𝑞 − 𝜗 𝑜 ∗2.4 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8= 𝜗 𝑞 − 𝜗 𝑜 ∗2.4 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 𝜗 𝑜 𝜗𝜗 𝜗 𝑜 𝑜𝑜 𝜗 𝑜 + 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 𝜗 𝑜 + 𝜗 𝑞 ∗1.8= 𝜗 𝑞 − 𝜗 𝑜 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 − 𝜗 𝑜 𝜗𝜗 𝜗 𝑜 𝑜𝑜 𝜗 𝑜 𝜗 𝑞 − 𝜗 𝑜 ∗2.4 𝜗 𝑜 + 𝜗 𝑞 ∗1.8= 𝜗 𝑞 − 𝜗 𝑜 ∗2.4 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8= 𝜗 𝑞 − 𝜗 𝑜 ∗2.4 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑜 + 𝜗 𝑞 ∗1.8= 𝜗 𝑞 − 𝜗 𝑜 ∗2.4 𝜗 0 𝑡 𝑥 =𝑆

1.8𝜗 0 +1.8 𝜗 𝑞 =2.4 𝜗 𝑞 −2.4 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 1.8𝜗 0 +1.8 𝜗 𝑞 =2.4 𝜗 𝑞 −2.4 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 1.8𝜗 0 1.8𝜗𝜗 1.8𝜗 0 0 1.8𝜗 0 +1.8 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 =2.4 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 −2.4 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 1.8𝜗 0 +1.8 𝜗 𝑞 =2.4 𝜗 𝑞 −2.4 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 1.8𝜗 0 +1.8 𝜗 𝑞 =2.4 𝜗 𝑞 −2.4 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 1.8𝜗 0 +1.8 𝜗 𝑞 =2.4 𝜗 𝑞 −2.4 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 1.4 𝜗 𝑜 =0.2 𝜗 𝑞 𝜗 0 𝑡 𝑥 =𝑆 1.4 𝜗 𝑜 =0.2 𝜗 𝑞 𝜗 0 𝑡 𝑥 =𝑆 1.4 𝜗 𝑜 𝜗𝜗 𝜗 𝑜 𝑜𝑜 𝜗 𝑜 =0.2 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 1.4 𝜗 𝑜 =0.2 𝜗 𝑞 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 1.4 𝜗 𝑜 =0.2 𝜗 𝑞 𝜗 0 𝑡 𝑥 =𝑆 1.4 𝜗 𝑜 =0.2 𝜗 𝑞 𝜗 0 𝑡 𝑥 =𝑆

𝜗 𝑞 =7 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑞 =7 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑞 𝜗𝜗 𝜗 𝑞 𝑞𝑞 𝜗 𝑞 =7 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝜗 𝑞 =7 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 𝜗 𝑞 =7 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 𝜗 𝑞 =7 𝜗 0 𝜗 0 𝑡 𝑥 =𝑆 7 𝜗 0 − 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 7 𝜗 0 − 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 7 𝜗 0 − 𝜗 0 7 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 − 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 7 𝜗 0 − 𝜗 0 ∗2.4=𝑆𝑆 7 𝜗 0 − 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 7 𝜗 0 − 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 7 𝜗 0 − 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 6 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 6 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 6 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 ∗2.4=𝑆𝑆 6 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =𝑆𝑆 6 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆 6 𝜗 0 ∗2.4=𝑆 𝜗 0 𝑡 𝑥 =𝑆

6 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 ∗2.4= 𝜗 0 𝜗𝜗 𝜗 0 0 𝜗 0 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 ; 𝑡 𝑥 𝑡𝑡 𝑡 𝑥 𝑥𝑥 𝑡 𝑥 =14,4 𝑠𝑠𝑜𝑜𝑎𝑎𝑡𝑡