Предмет: Физика, автор: kristinachermenyova

Определи радиус выпуклого

моста, если при движении автомобиля со скоростью 72 км/ч

в верхней точке сила давления на

мост в 2 раза меньше силы

тяжести автомобиля.

Ответы

Автор ответа: elena20092

0

Ответ:

Радиус моста 80 м

Объяснение:

v = 72 км/ч = 20 м/с

g = 10 м/с² - ускорение свободного падения

P = 0.5 mg (m - масса автомобиля)

R - ?

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В верхней точке моста на автомобиль действует со стороны моста сила N = P = 0,5 mg, направленная вверх, и сила тяжести mg, направленная вниз. При этом автомобиль имеет центростремительное ускорение а = v²/R, также направленное вниз, к центру кривизны моста.

Согласно 2-му закону Ньютона

ma = mg - N

или

mv²/R = mg - 0.5 mg

v²/R = 0.5g

Откуда

R = 2v²/g = 2 · 20² : 10 = 80 (м)

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