a proton at the large hadron collider at CERN travels at 295,000,000 meters per second. How long does it take the proton to go around the 17 mile circumference of the collider?

9.27 x 10^-5 seconds

Step-by-step explanation:

We are given the speed of the proton at which it travels i. e. 295,000,000 meters per second.

And we also know the total circumference of the collider = 17 mile.

So to find the time taken the proton to travel around the circumference of the collider, we first need to make the units same.

1 mile = 1609.34 meters

17 mile = 1609.34 * 17 = 27358.78

Time taken by the proton = distance / speed

= 27358.78 / 295000000

= 9.27 x 10^-5

Therefore, it takes 9.27 x 10^-5 or 0.0000927 seconds for the proton to go around 17 mile circumference of the collider.

Answer: It takes 0.000092742 seconds the proton to go arond the 17 mile circumference of the collider.

Solution:

Velocity: v=295,000,000 meters / second

Time: t=?

Distance: d=17 miles (1609.344 meters / 1 mile) →d=27,358.848 meters

v=d/t

Replacing the given values:

295,000,000 meters / second = 27,358.848 meters / t

Solving for t:

(295,000,000 meters / second) t = 27,358.848 meters

t = (27,358.848 meters) / (295,000,000 meters / second)

t=0.000092742 seconds