A rectangular plate of glass, measured at rest, has sides \(30.0 \mathrm{cm}\) and \(60.0 \mathrm{cm} .\) (a) As measured in a reference frame moving parallel to the 60.0 -cm edge at speed \(0.25 c\) with respect to the glass, what are the lengths of the sides? (b) How fast would a reference frame have to move in the same direction so that the plate of glass viewed in that frame is square?

$$L = (60.0 \mathrm{cm})\sqrt{1-\frac{(0.25)^2}{1}}$$ $$L = (60.0 \mathrm{cm})\sqrt{1-0.0625}$$ $$L = (60.0 \mathrm{cm})\sqrt{0.9375}$$ $$L \approx 58.08 \mathrm{cm}$$ So the length of the 60.0 cm side in the moving frame is approximately 58.08 cm. #tag_title#Determining the speed for the plate's sides to appear equal#tag_content#Now, we must find the speed for the plate's sides to appear equal in a moving frame. In this case, we want the contracted length of the 60.0 cm side to be equal to the 30.0 cm side. Using the length contraction formula: $$30.0 \mathrm{cm} = (60.0 \mathrm{cm})\sqrt{1-\frac{v^2}{c^2}}$$ Solve for \(v\): $$\frac{30.0 \mathrm{cm}}{60.0 \mathrm{cm}} = \sqrt{1-\frac{v^2}{c^2}}$$ $$\frac{1}{2} = \sqrt{1-\frac{v^2}{c^2}}$$ Square both sides: $$\frac{1}{4} = 1-\frac{v^2}{c^2}$$ Move the fractions: $$\frac{v^2}{c^2} = 1 - \frac{1}{4}$$ $$\frac{v^2}{c^2} = \frac{3}{4}$$ Multiply by \(c^2\) and take the square root: $$v = c\sqrt{\frac{3}{4}}$$ $$v = c\sqrt{0.75}$$ Therefore, the speed required for the plate's sides to appear equal in a moving frame is approximately 0.866 times the speed of light, or 0.866c.