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fredd [130]
3 years ago
9

Do two sound waves that seen equally loud always have the same amplitude? explain

Physics
1 answer:
givi [52]3 years ago
3 0
No, because of that "seem" part of the question.

Your hearing system is not a linear sound detector and meter. 
Sounds with equal amplitudes at different frequencies "seem"
to have different loudness, and the differences are different
for different people.

You've probably heard the saying "Things are not always as they seem."
The fact is, they NEVER are.
You might be interested in
Part A: Determine the wavelength of photons that can be emitted
torisob [31]

Answer:

A  λ = 97.23 nm

, B)   λ = 486.2 nm

, C)  λ = 53326 nm

Explanation:

With that problem let's use the Bohr model equation for the hydrogen atom

          E_{n} = -k e² /2a₀  1/n²

For a transition between two states we have

          E_{nf} -  E_{no} = -k e² /2a₀ (1/  n_{f}² - 1 / n₀²)

Now this energy is given by the Planck equation

         E = h f

And the speed of light is

         c = λ f

Let's replace

      h c / λ = - k e² /2a₀ (1 / n_{f}² - 1 / no₀²)

      1 /  λ = - k e² /2a₀ hc (1 / n_{f}² -1 / n₀²)

Where the constants are the Rydberg constant R_{H} = 1.097 10⁷ m⁻¹

        1 /  λ = R_{H} (1 / n₀² - 1 / nf²)

Now we can substitute the given values

Part A

 Initial state n₀ = 1 to the final state n_{f} = 4

        1 /  λ = 1.097 10⁷ (1/1 - 1/4²)

         1 /  λ = 1.0284 10⁷ m⁻¹

          λ = 9.723 10⁻⁸ m

We reduce to nm

         λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)

         λ = 97.23 nm

Part B

Initial state n₀ = 2 final staten_{f} = 4

       1 /  λ = 1.097 10⁷ (1/2² - 1/4²)

       1 /  λ = 0.2056 10⁻⁷ m

        λ = 486.2 nm

Part C

Initial state n₀ = 3

      1 /  λ = 1,097 10⁷ (1/3² - 1/4²)

       1 /  λ = 5.3326 10⁵ m⁻¹

        λ = 5.3326 10-5 m

        λ = 53326 nm

5 0
3 years ago
The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t
oee [108]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

PE=\frac{GMm}{r}

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

Then the potential energy at this point would be,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

The we calculate the Potential gravitational energy,

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

7 0
3 years ago
A rocket traveling at 150 m/s is accelerated for 15 seconds at a rate of 4.5
Dmitrij [34]

Answer:

a) vf = vi + at

Explanation:

a) vf = vi + at

7 0
3 years ago
A child with a weight of 110 N swings on a playground swing attached to 2.00 m long chains. What is the gravitational potential
Ierofanga [76]

Answer

given,

Weight of the child = 110 N

length of the swing,L = 2 m

now, calculating the potential energy when the string is horizontal

  Potential energy = m g h

 now, h = L (1 - cos θ)   where θ is the angle made by the string with the vertical.

     PE = m g L (1 - cos θ)

   when rope is horizontal θ = 90°

     PE = 110 x 2 (1 - cos 90°)

    PE = 220 J

now, calculating potential energy when string made 25° with horizontal

PE = m g L (1 - cos θ)

   when rope is horizontal θ = 25°

     PE = 110 x 2 (1 - cos 25°)

     PE = 20.61 J

5 0
4 years ago
An object moving north with an initial velocity of 14m/s accelerates 5m/s squared. What is the final velocity of the object?
OLEGan [10]
The final velocity of the object is 16m\s.
Hope this helps! :)
6 0
3 years ago
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