One of the limiting factors in high-performance aircraft is the acceleration to which the pilot can be subjected without blacking out; it’s measured in “gees,” or multiples of the gravitational acceleration. The F-22 Raptor fighter can achieve Mach 1.8 (1.8 times the speed of sound, which is about 340 m/s). Suppose a pilot dives in a circle and pulls up.
If the pilot can’t exceed 6.0 g, what’s the tightest circle (smallest radius) in which the plane can turn?
R_min= _____ m
If the pilot can’t exceed 6.0 g, what’s the tightest circle (smallest radius) in which the plane can turn?
R_min= _____ m
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Skidude is almost right, but the question ambiguously telling us the *speed of sound* is 340m/s (which it roughly is), not that Mach 1.8 = 340m/s. The actual velocity is 1.8*340 = 612m/s.
So the formula becomes,
6.0(9.8m/s^2)=(612m/s)^2/r
So the formula becomes,
6.0(9.8m/s^2)=(612m/s)^2/r
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Rearrange the equation by multiplying both sides by r, then dividing both sides by 6*9.8, which gives,
r = (612^2)/(6*9.8)
r = (612^2)/(6*9.8)
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ok, so given the nature of the problem I'm assuming you can do the math if I show you the concepts. so.....
centripetal acceleration= (speed)^2/(radius)
We know that the maximum acceleration is 6.0g which can also be seen as 6.0(9.8m/s^2)=whatever. Also, we know the speed to be 340m/s. Now just substitute those values in to the above equation.
6.0(9.8m/s^2)=(340m/s)^2/r
solve for r and that should be the right answer!
centripetal acceleration= (speed)^2/(radius)
We know that the maximum acceleration is 6.0g which can also be seen as 6.0(9.8m/s^2)=whatever. Also, we know the speed to be 340m/s. Now just substitute those values in to the above equation.
6.0(9.8m/s^2)=(340m/s)^2/r
solve for r and that should be the right answer!