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Taya2010 [7]
3 years ago
6

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

) What is the pressure amplitude of the sound?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

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Which of the following is NOT a potential result of climate change?
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Answer:

My answer :

Explanation:

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When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a
Nana76 [90]

Answer:

  Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

      Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

       Фi = ∫ E .dA = q_{int} / ε₀

the Gaussian surface is a sphere so its area is

        A = 4 π r²

the charge inside is

        q_{int} = Q

we substitute

       E 4π r² = Q /ε₀

       E = 1 / 4πε₀   Q / r²

To calculate the flow on the two surfaces

* Sphere

       Ф = E A

        Ф = 1 / 4πε₀  Q / r² (4π r²)

        Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

          r = d / 2

We look for the diagonal with the Pythagorean theorem

         d² = L² + L² = 2 L²

         d = √2 L

         

we substitute

          r = √2 / 2 L

          r = L / √2

          L = √2  r

now we can calculate the area of ​​the cube that has 6 faces

          A = 6 L²

          A = 6 (√2  r)²

          A = 12 r²

the flow is

          Ф = E A

          Ф = 1 / 4πε₀  Q/r²  (12r²)

          Ф_cubo = 3 /πε₀  Q

the relationship of these two flows is

         Ф_cube /Ф_sphere = 3 /π

8 0
3 years ago
An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other
Umnica [9.8K]
The answer is c. +2.0 µC

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹  N*m/C²


It is given:

F = 7.2 N

d = 0.1 m =  10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Q2 = ?


Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²

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Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶ 


Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

5 0
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