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eduard
3 years ago
15

A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wh

eel (nose wheel) is 0.800 m behind the nose, and the main wheels are 3.02 m behind the nose. What percentage of the airplane's weight is supported by the nose wheel?
Physics
1 answer:
otez555 [7]3 years ago
7 0

Answer:

The percentage of the weight supported by the front wheel is  A= 19.82 %

Explanation:

From the question we are told that

   The center of gravity of the plane to its nose  is  z = 2.58 m

    The distance of the front wheel of the plane to  its nose is l = 0.800\ m

     The distance of the main wheel of the plane to its nose is e =  3.02 \ m

At equilibrium  the Torque about the nose of the airplane is mathematically represented as

          mg (z- l) -  G_B *(e - l) = 0

Where m is the mass of the airplane

          G_B is the weight of the airplane supported by the main wheel  

       So  

             G_B =\frac{mg (z-l)}{(e - l)}

Substituting values

            G_B =\frac{mg (2.58 -0.8 )}{(3.02  - 0.80)}

           G_B = 0.8018 mg

Now the weight supported at the frontal wheel is mathematically evaluated as

           G_F = mg - G_B

Substituting values      

       G_F = mg - 0.8018mg    

      G_F = (1 - 0.8018) mg      

     G_F = 0.1982 mg    

Now the weight of the airplane is  =  mg

Thus percentage of this weight supported by the front wheel is  A = 0. 1982 *100 = 19.82 %

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Option C is the untrue statement.

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a_E=\dfrac{Gm}{r^2}

Explanation:

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The force on the person will balance the gravitational force

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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

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The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

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(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

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