JAMB - Physics (1985 - No. 11)
A force of 100N stretches an elastic string to a total length of 20cm. If an additional force of 100N stretches the string 5cm further, Find the natural length of the string
15cm
12cm
10cm
8cm
5cm
Giải thích
Using, f1 = f2
e1 = 20 - L ; e2 = 5cm
\(\frac{f_1}{e_1}\) = \(\frac{f_2}{e_2}\)
\(\frac{100}{20 - L}\) = \(\frac{100}{5}\)
hence, L = 15cm
e1 = 20 - L ; e2 = 5cm
\(\frac{f_1}{e_1}\) = \(\frac{f_2}{e_2}\)
\(\frac{100}{20 - L}\) = \(\frac{100}{5}\)
hence, L = 15cm