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Crazy boy [7]
4 years ago
9

An F-15 is flying straight and level lift equals weight) at 8 km standard day conditions at a velocity of 260 m/s. Given the fol

lowing data for the F-15, answer the questions that follow: Parasite/Profile drag coefficient CD0-0.018, wing span b-12.9 m, wing area s = 56.7 m2, mass m = 22.680 kg. Oswald efficiency factor eo0.915
(a) What is the wing's aspect ratio A?
(b) What is the lift force L generated by the aircraft?
(c) What is the aircraft's drag coefficient CD?
(d) What is the drag force D generated by the aircraft?

Physics
1 answer:
topjm [15]4 years ago
5 0

Answer:

1)Wing ratio=2.935

2)Lift force=l=222264 N

3)Drag coefficient=Cd=0.0237

4)Drag force generated by aircraft=Cd=2388.189 N

Explanation:

Solution explanation is given in attachments.

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Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

v_{i}=-44.5m/s\\v_{f}=55.5m/s

Now we need to find the initial momentum

So

P_{1}=m*v_{i}

Substitute the given values

P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s

Now for final momentum

P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s

So the change in momentum is given as:

ΔP=P₂-P₁

=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s

ΔP=20 kg.m/s

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