# Chapter 26 - Problems - Page 1010: 20

(a) As measured in the moving reference frame, the lengths of the sides are 30 cm and 58 cm (b) $v = 0.866~c$ $v = 2.6\times 10^8~m/s$

#### Work Step by Step

(a) Let $L_0 = 60~cm$ We can find the length $L$ according to the moving reference frame: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $L = (60~cm)~\sqrt{1-\frac{(0.25~c)^2}{c^2}}$ $L = (60~cm)~\sqrt{1-\frac{1}{16}}$ $L =58~cm$ Since the 30-cm side is perpendicular to the direction of motion, there is no length contraction of this side. As measured in the moving reference frame, the lengths of the sides are 30 cm and 58 cm (b) Let $L_0 = 60~cm$ and let $L = 30~cm$ We can find the required speed $v$ of a moving reference frame: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{L}{L_0}$ $1-\frac{v^2}{c^2} = (\frac{L}{L_0})^2$ $\frac{v^2}{c^2} = 1-(\frac{L}{L_0})^2$ $v^2 = [1-(\frac{L}{L_0})^2]~(c^2)$ $v = \sqrt{1-(\frac{L}{L_0})^2}~\times c$ $v = \sqrt{1-(\frac{30~cm}{60~cm})^2}~\times c$ $v = 0.866~c$ $v = (0.866)~(3.0\times 10^8~m/s)$ $v = 2.6\times 10^8~m/s$

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