A stunt pilot of mass 55.0 kg who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane.
QUESTION 1:
If the plane's speed at the lowest point of the circle is 95.8 m/s , what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00g ?
Take the free fall acceleration to be g= 9.80 m/s^2.
ANSWER IN METERS
QUESTION 2:
What is the apparent weight of the pilot at the lowest point of the pullout in this case?
Take the free fall acceleration to be g= 9.80 m/s^2 .
ANSWER IN NEWTONS
2007-03-19 23:36:24 · 1 answers · asked by Mr_ 1 in Science & Mathematics ➔ Physics
1. Acceleration of the uniform rotational movement is a = v^2/R.
If your acceleration is maximum 4g then your minimum R = v^2/4g = 234 m.
2. Pilot experiences 4g + g = 5g acceleration. Since the weight (force) is m(5g) his weight will be 2695 N or 269.5 kg*.
2007-03-20 02:39:14 · answer #1 · answered by fernando_007 6 · 1⤊ 0⤋