Question

A roller coaster has a loop in which the centripetal acceleration equals 9.8m/s2 . Of the tangential speed of the roller coaster cars is 15-7 m/s

Answer 1

The question involves calculating the speed of a roller coaster at the top of a loop required to maintain a centripetal acceleration of 1.50 times the acceleration due to gravity, using the formula a = v^2/r for centripetal acceleration.

Step-by-step explanation:

The question pertains to the concept of centripetal acceleration in physics. Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. For an object moving with speed v along a path with a radius of curvature r, its centripetal acceleration a can be calculated using the formula a = v^2/r.

In the scenario described by the question, we are asked to find the speed of a roller coaster at the top of a loop where the centripetal acceleration is 1.50 times the acceleration due to gravity (g), which is 9.8 m/s2. Given that 1.50 g is equal to 1.50 x 9.8 m/s2, we can set up the equation a = v^2/r and solve for v to find the speed required to achieve this centripetal acceleration at the top of the loop with a radius of 15 m.

Answer 2

25.152m

Step-by-step explanation:

Compete question;

A roller coaster has a loop in which the centripetal acceleration equals 9.8m/s² . If the tangential speed of the roller coaster cars is 15.7 m/s, find its radius.

centripetal acceleration a = v²/r

v is the tangential speed

r is the radius

Given

v = 15.7m/s

a = 9.8m/s²

9.8 = 15.7²/r

9.8r = 15.7²

r = 15.7²/9.8

r = 246.49/9.8

r = 25.152m

Hence the radius of the coaster is 25.152m