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Artemon [7]
4 years ago
11

An airplane has been loaded in such a manner that the CG is located aft of the aft CG limit. One undesirable flight characterist

ic a pilot might experience with this airplane would be ______.a. a longer takeoff run.
b. difficulty in recovering from a stalled condition.
c. stalling at higher-than-normal airspeed.
Physics
1 answer:
RoseWind [281]4 years ago
4 0

Answer:

Option B is correct ( difficulty in recovering from a stalled condition.)

Explanation:

Option B is correct ( difficulty in recovering from a stalled condition.)

If you load airplane in such manners CG has changed then it will have huge impact on the longitudinal stability. Due to this airplane would face  difficulty in recovering from a stalled condition.

Stall is the phenomena in which air resistance is increased on the wings which causes decrease in lift. So it is important to remain with in CG limit put by manufacturer to avoid stall condition.

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a. Sweet corn and possibly d. okra.

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4 years ago
One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.
Nadya [2.5K]
The original Coulomb force between the charges is:

Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.

The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive. 

Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
Q₂ is decreased to Q₂₂=(1/2)*Q₂=Q₂/2. 

New force:

Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge

Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:

Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,

Fc₁=(k*(1/6)*Q₁*Q₂)/r²

Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:

Fc₁=(1/6)*Fc

So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc. 
4 0
3 years ago
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A solid plastic ball hits a stationary tennis ball in a perfectly elastic collision. If the mass of the plastic ball is three ti
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Tennis ball will move with 1/3 th velocity of plastic ball
8 0
4 years ago
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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim
Svetach [21]

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

a_c=\dfrac{V^2}{R}

Now by putting the values in the above equation we get

a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

7 0
4 years ago
What is the potential energy of a falling rock if it has a mass of 64 kg and fell from a height of 54 m
Talja [164]

<u>P.E (Potential Energy) = 33868.8 J</u>

<u>Explanation:</u>

The formula to calculate Potential energy is:

PE = mgh

where PE = potential energy

m = mass of an object

g = acceleration due to gravity

h = height of the object

Here, m = 64 kg

         g  =  9.8 m / s^{2}

         h  = 54 m

So, substituting the values in the formula:

PE = mgh

    = 64 × 9.8 × 54

    = <u> 33868.8 J</u>

6 0
3 years ago
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