## An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then travels at n

Question

An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then travels at near light speeds parallel to the length (L) of the box. If the observer measures the width (w) of the box while moving, the measured value will now be:________

a. equal to w.
b. less than w.
c. greater than w.
d. information is needed. You need to know the density of the box.

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1 year 2021-09-05T03:46:30+00:00 1 Answers 17 views 0

b. less than w.

Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

L = $$L_0\sqrt{1-\frac{v^2}{c^2} }$$

Here, L is the observed length

$$L_0$$ is the original length of the object

v is the relative speed between the object and the observer

and c is the speed of light

Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

This effectively decreases the length that is observed.