Предмет: Физика, автор: olga88000

ПОЖАЛУЙСТА,ПОМОГИТЕ!ЗАВТРА КОНТРОЛЬНАЯЯЯ

Определите амплитуду гармонических колебаний тела, если его полная энергия 0,04 Дж, а жесткость пружины, к которой прикреплено тело, равна 2 Н/м.

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Поплавок на воде совершает колебания с частотой 4 Гц. Какова скорость волн, разбегающихся от поплавка по поверхности воды, если расстояние между их гребнями составляет 1,5 см?

Ответы

Автор ответа: Okita

1 задача

E=(kA^2)/2

A^2=(2E)/k =(2*0,04)/2=0,04

тогда A=0,2

2 задача

1,5 cм= 1,5*10^-2

V(скорость)=л(длина волны)*v(частота волны)

V=4*1,5*10^-2=6*10^-2=0,06

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