Предмет: Физика, автор: sanowersholki7

ПОЖАЛУЙСТА ПОМОГИТЕ!!!
линейная скорость точки, находящейся на экваторе Земли, равна 465м/с.Радиус Земли 6400км
а)вычеслите угловую скорость движения этой точки;
б) определите центростремительное ускорение точки на экваторе;

колесо велосипеда имеет радиус 40см с какой скоростью едет велосипедист, если колесо делает 120об/мин? чему равна его центростремительное ускорение.

Ответы

Автор ответа: zinovicgrigorij
1

Ответ: 7,26*10⁻⁵рад/с; 0,03м/с²; 5,024м/с; 63,1м/с²

Объяснение:

Дано:

V=465м/с

R=6400км=6400000м

-------------------

ω-?; а-?

ω=V/R=465/6400000=7,26*10⁻⁵рад/с

а=V²/R=465²/6400000=0,03м/с²

Дано:

R=40см=0,4м

n=120об/мин

------------------

V-?; а-?

V=ω*R=π*n*R/30=3,14*120*0,4/30=5,024м/с

а=V²/R=5,024²/0,4=63,1м/с²


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