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Vadim26 [7]
4 years ago
6

A jet with mass m = 90000.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.

Physics
1 answer:
leva [86]4 years ago
6 0
This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)
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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

     T_2 = 1.008 % higher than T_1

    T_2 = 0.99 % lower than T_1

Explanation:

   From the question we are told that

         The first string has a frequency of   f_1 = 230 Hz

          The period of the beat is  t_{beat}  = 0.99s

Generally the frequency of the beat is

             f_{beat} = \frac{1}{t_{beat}}

  Substituting values

            f_{beat} = \frac{1}{0.99}

                   = 1.01 Hz

From the question

        f_2 - f_1 = f_{beat}   for  f_2  having a  higher tension

So

       f_2 - 230 = 1.01

               f_2 = 231.01Hz

 From the question

            \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }

         \frac{T_2}{T_1}  = \frac{f_2^2}{f_1^2}

Substituting values

         \frac{T_2}{T_1}  = \frac{(231.01)^2}{(230)^2}

      T_2 = 1.008 % higher than T_1

    For f_2 having a lower tension

           f_1 - f_2 = f_{beat}

  So

       230 - f_2 = 1.01

            f_2 = 230 -1.01

                  = 228.99

  From the question

            \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }

         \frac{T_2}{T_1}  = \frac{f_2^2}{f_1^2}

    Substituting values

         \frac{T_2}{T_1}  = \frac{(228.99)^2}{(230)^2}

      T_2 = 0.99 % lower than T_1        

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