Answer
11 cubic meters
Work Step by Step
The side of the box parallel to the direction of motion is contracted. Use equation 26–3a.
$$\mathcal{l} = \mathcal{l}_{o} \sqrt{1-\frac{ v^{2} } { c^{2} }} $$
$$= (2.6m) \sqrt{1-\frac{(0.80c)^2} { (c)^{2} }} =1.56m$$
The lengths of the other sides that are perpendicular to the motion are unchanged. Find the volume.
$$V=(1.56m)(2.6m)^2=10.55m^3\approx 11m^3$$
