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Setler [38]
3 years ago
9

The blackbox recorder from an airplane that hit a patch of severe turbulence indicated that the plane moved up and down with an

amplitude of 30.0 m and a maximum acceleration of 1.8 g. Assuming the vertical motion was SHM Find the period Find the plane’s maximum vertical speed
Physics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

T = 8.19 s ,  v_{max}  = 23 m / s

Explanation:

In the simple harmonic motion the equation that describes them is

    y = A cos wt

Acceleration can be found by derivatives

    a = d²y / dt²

   v = dy / dt = - Aw sin wt

   a= d²y / dt² = - A w² cos wt

For maximum acceleration cosWT = + -1

    a_{max} = -A w2

    w = RA (a_{max}/ A)

    w = RA (1.8 9.8 / 30.0)

    w = 0.767 rad / s

The angular velocity is related to the frequency

    w = 2π f

     f = 1 / T

    w = 2π / T

    T = 2π / w

    T = 2π / 0.767

    T = 8.19 s

For maximum speed the sin wt = + -1

    v_{max} = A w

    v_{max}  = 30.0 0.767

   v_{max}  = 23 m / s

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3 years ago
What is the wavelength of an fm radio signal that has a frequency of 91.7 mhz?
Digiron [165]

Answer:

The wavelength of the radio wave will be 3.271 m

Explanation:

We have given frequency of the radio signal f=91.7MhZ=91.7\times 10^6Hz

Speed of the light c=3\times 10^8m/sec

We have to find the wavelength of the radio signal

We know that wavelength is given by \lambda =\frac{c}{f}=\frac{3\times 10^8}{91.7\times 10^6}=3.271m

So the wavelength of the radio wave will be 3.271 m

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Answer:

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Explanation:

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Westkost [7]

Answer:

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Explanation:

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nikitadnepr [17]

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