Question

10. Luther is designing a roller coaster for an amusement park. At one point, the roller coaster will enter a horizontal loop at the speed of 31.8 m/s. If Luther does not want to centripetal acceleration to exceed 29.0 m/s^2, what is the minimum radius of the horizontal loop?

16. When a certain roller coaster is at the top of a 61 m hill, it has no velocity. As the coaster descends the hill, it gains speed. What is the roller coasters speed when the it has a height of 34 m above the ground?

Please answer both of you can!!

Answer 1

The minimum radius of the horizontal loop on the roller coaster to keep the centripetal acceleration below 29.0 m/s^2 is calculated using the formula r = v^2/a, yielding approximately 34.979 meters.

To calculate the minimum radius of the horizontal loop so that the centripetal acceleration does not exceed 29.0 m/s2, we use the formula for centripetal acceleration, which is a = v2/r, where a is the centripetal acceleration, v is the velocity, and r is the radius of the loop. Rearranging the formula to solve for r, we get r = v2/a. Substituting the given values, we calculate the minimum radius as r = (31.8 m/s)2 / 29.0 m/s2, which equals approximately 34.979 meters.

Therefore, the minimum radius of the horizontal loop should be about 34.979 meters to ensure that the centripetal acceleration does not exceed 29.0 m/s2.

Answer 2

centripital acceleration= v^2/r

r = v^2/a

r=31.8×31.8/29

r=34.8703m

Step-by-step explanation:

the maximum acceleration is obtained with minimum radius.