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shusha [124]
3 years ago
15

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

intensity of wave 1 to the intensity of wave 2?
Physics
1 answer:
zlopas [31]3 years ago
8 0

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

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An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a
SCORPION-xisa [38]

Answer:

a_y=0.92m/s^2

Explanation:

To solve this problem we use the formula for accelerated motion:

y=y_0+v_{y0}t+\frac{a_yt^2}{2}

We will take the initial position as our reference (y_0=0m) and the downward direction as positive. Since the rock departs from rest we have:

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Using our values:

a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2

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Urgent if earths gravitational field strength is 10 N/kg how many newtons does a 30kg object weigh
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a spring has a force constant of 100 n/m and an unstretched length of 0.07 m. one end is attached to a post that is free to rota
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F equals 3N with respect to the circle's center, moving in the same direction as the centripetal acceleration.

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1 year ago
What is the IMA of the following pulley system?<br><br>34567
Lynna [10]

Answer:

    IMA of given system =   \frac{F_{r} }{F_{e} }

Explanation:

  • The "Ideal Mechanical advantage" (IMA) of given pulley is \frac{F_{r} }{F_{e} } .
  • Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by F_{r}. The input force is analogous or equivalent to the effort applied i.e. F_{e} .
  • Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is  \frac{F_{r} }{F_{e} } .
  • Its mathematical reference would be:

                                                IMA =   \frac{F_{r} }{F_{e} }

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