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Sedaia [141]
3 years ago
6

Just need to confirm an answer

Physics
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

B

Explanation:

Remark

If, when it reaches its maximum height where the PE is the greatest, that the maximum energy is when the ball is first kicked. That would mean its starting energy is 10 Joules.

Formula

KE = 1/2 m v^2

Givens

KE = 10 J

m = 0.2 kg

v = ?

Solution

10 = 1/2 * 0.2 * v^2

10 * 2 = 0.2 * v^2

20 = 0.2 * v^2

20/0.2 = v^2

100 = v^2

v = 10 m/s

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Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out
PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is  

\lambda _2 = (6.48,6.52) \ cm

Explanation:

From the question we are told that

     The wavelength of the note is \lambda  =  6.50 \ cm = 0.065 \ m

     The difference in beat frequency is \Delta  f = 17.0 \ Hz

     

Generally the frequency of the note played by the guitar that is in tune is  

        f_1 = \frac{v_s}{\lambda}

Where v_s is the speed of sound with a constant value v_s  =  343 \ m/s

       f_1 = \frac{343}{0.0065}

      f_1 = 5276.9 \ Hz

The difference in beat is mathematically represented as

       \Delta  f =  |f_1 - f_2|

Where f_2 is the frequency of the sound from the out of tune guitar

     f_2 =f_1  \pm \Delta f

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      f_2 = 5276.9 + 17.0  

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The wavelength for this frequency is

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     \lambda_2 = 0.0648 \ m

    \lambda_2 = 6.48 \ cm

For the second value of the second frequency

     f_2 =  f_1 - \Delta f

     f_2 = 5276.9 -17

      f_2 = 5259.9 Hz

The wavelength for this frequency is

   \lambda _2 = \frac{343}{5259.9}

   \lambda _2 = 0.0652 \ m

   \lambda _2 = 6.52 \ cm

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