A jet pilot is in a vertical loop. If the jet is moving at a speed of 1300 km per hour at the lowest point of the loop determine the minimum radius of the cirlce so that the centripetal acceleration at the lowest point does not exceed 6.0g's.

6*9.8m/s^2>mg+mv^2/r solve for r.

change v to m/s

To find the minimum radius of the circle, we need to consider the centripetal acceleration at the lowest point of the loop.

Centripetal acceleration is given by the formula:

ac = (v^2) / r

Where:
- ac is the centripetal acceleration
- v is the speed of the jet
- r is the radius of the circle

In this case, the centripetal acceleration should not exceed 6.0g's. One "g" is equal to the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, 6.0g's can be represented as 6.0 * 9.8 m/s^2.

Converting the speed from km/h to m/s:
1300 km/h = (1300 * 1000) / (60 * 60) m/s = 361.1 m/s

Now, let's substitute the given values into the formula:

6.0g's = 6.0 * 9.8 m/s^2 = 58.8 m/s^2
v = 361.1 m/s
ac = 58.8 m/s^2

58.8 m/s^2 = (361.1 m/s)^2 / r

Simplifying the equation:
r = (361.1 m/s)^2 / 58.8 m/s^2

Calculating:
r = (130441.21 m^2/s^2) / 58.8 m/s^2
r ≈ 2213.9 meters

Therefore, the minimum radius of the circle should be approximately 2213.9 meters for the centripetal acceleration at the lowest point to not exceed 6.0g's.