DIFFICULT PHYSICS PROBLEM! Help!!!
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DIFFICULT PHYSICS PROBLEM! Help!!!

[From: ] [author: ] [Date: 11-10-08] [Hit: ]
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-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.......
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

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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

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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)

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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!
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