English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

What is its amplitude to the nearest centimeter?

2007-12-17 11:29:27 · 2 answers · asked by Anonymous in Science & Mathematics Engineering

2 answers

if the amplitude is the maximum (lateral) distance from the centre of oscillations then the answer is just
200*sin(24.4 degrees) = 82.62 cm
to the nearest cm that's 83 cm.

the mass appears to be irrelevant in this question

2007-12-19 13:59:33 · answer #1 · answered by Anonymous · 0 0

A = 200sin(24.4) ≈ 82.62089 cm ≈ 83 cm
The half-swing is 200π(24.4)/180 ≈ 85 cm

2007-12-19 14:06:29 · answer #2 · answered by Helmut 7 · 1 0

fedest.com, questions and answers