Answer:
[tex]V_o=25.725\ m^3[/tex]
Explanation:
Given:
sides of the cube, [tex]a=3.5\ m[/tex]
speed of the cube with respect to the observer, [tex]v=0.8c[/tex]
Since the relative velocity of the object is relativistic, so there will be a length contraction according to the observer:
[tex]a_o=a\div\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]
where:
[tex]a_o=[/tex] observed length of the side along the direction of velocity
[tex]a_o=3.5\div\frac{1}{\sqrt{1-\frac{(0.8c)^2}{c^2} } }[/tex]
[tex]a_o=2.1\ m[/tex] is the observed length of the cube edge only in the direction of the velocity due to relativistic effect of length contraction.
So the observed volume will be:
[tex]V_o=a\times a\times a_o[/tex]
[tex]V_o=3.5\times 3.5\times 2.1[/tex]
[tex]V_o=25.725\ m^3[/tex]
The volume of the box observed at the given speed is 25.73 m³.
The given parameters;
The edge length of the cube according to the stationary observer is calculated as;
[tex]a' = a\times \sqrt{1- \frac{v^2}{c^2} } \\\\a' = 3.5 \times \sqrt{1 - \frac{(0.8c)^2}{c^2} }\\\\a' = 3.5 \times 0.6\\\\a' = 2.1 \ m[/tex]
The volume of the cube is calculated as follows;
[tex]V = A \times a'\\\\V = (a^2) \times a'\\\\V = (3.5)^2 \times 2.1\\\\V = 25.73 \ m^3[/tex]
Thus, the volume of the box observed at the given speed is 25.73 m³.
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