91影视

We have given that, A cosmic ray travels $60km$through the earth鈥檚 atmosphere in $400渭s,$as measured by experimenters on the ground.

The cosmic ray enters the atmosphere and the cosmic ray hits the ground are two separate events.

In the cosmic ray's frame, frame$S^{\text{'}}$these can both be measured with a single clock. As a result, in $S^{\text{'}}$the proper time interval localid="1649596902268" $鈭�t^{\text{'}}=鈭�蟿$exists between them. The time interval measured in the Earth's frame $S,$is $鈭�t=400渭s.$

The following is the outcome of time dilation:

$鈭�蟿=\sqrt{1-{尾}^2}鈭�t$

The cosmic ray's speed in frame $S$is simply:

$$\begin{gathered} v=\frac{鈭�L}{鈭�t} \\ v=\frac{60000}{4.00脳{10}^{-6}} \\ v=1.5脳{10}^{8}m/s \end{gathered}$$

From previous, we have to find value for $尾:$

$$\begin{gathered} 尾=\frac{v}{c} \\ 尾=\frac{1.5脳{10}^{8}m/s}{3脳{10}^{8}m/s} \\ 尾=0.5 \end{gathered}$$

Thus, the time interval measured by cosmic ray is:

$$\begin{gathered} 鈭�蟿=\sqrt{1-{(0.5)}^2}(400渭s) \\ 鈭�蟿=346渭s \end{gathered}$$