An airplane is flying in a horizontal circle at a speed of 104 m/s. The 88.0 kg pilot does not want the centripetal acceleration to exceed 6.15 times free-fall acceleration.
Find the minimum radius of the plane’s
path. The acceleration due to gravity is 9.81m/s2.
Answer in units of m
Dave your dealing with Circular motion so you would use this circular motion formula;Ac=Vc2/r
1. In your problem your trying to find the radius.
2. The Ac is already given, but since the problem asks for it not to exceed 6.15 times the free-fall acceleration. You must times them.
Ac=6.15*9.81
Ac=~60.3315m/s.
3. Next, you plug in your airplane speed (104m/s.) and square that.
4. Your final step is to divide 104(squared) by 6.15*9.81.
Giving you the answer of ~179.276m.
Good Luck! With the rest of your Physics!
To find the minimum radius of the airplane's path, we need to use the formula for centripetal acceleration:
a = v^2 / r
Where:
a is the centripetal acceleration,
v is the velocity of the airplane, and
r is the radius of the path.
Given information:
v = 104 m/s
a = 6.15 * g
g is the acceleration due to gravity, which is 9.81 m/s^2.
Substituting the values into the formula, we have:
6.15 * g = v^2 / r
Rearranging the equation to isolate r, we get:
r = v^2 / (6.15 * g)
Now we can substitute the values:
r = (104^2) / (6.15 * 9.81)
Now we can calculate the value of r:
r ≈ 1878.2 m
Therefore, the minimum radius of the plane's path is approximately 1878.2 m.