Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy
2 answers:
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The distance of the sound from the tuba is 4.82 m.
<h3>Area of the tube</h3>
The area of the tuba is calculated as follows;
I = P/A
where;
- I is intensity of sound
- P is power
- A is area
A = P/I
A = 0.35 / (1.2 x 10⁻³)
A = 291.67 m²
<h3>Distance of the sound</h3>Area = 4πr²
Thus, the distance of the sound from the tuba is 4.82 m.
Learn more about intensity of sound here: brainly.com/question/4431819
Answer:
The force magnitude is 1.75 μN acting outward from the origin towards x₂
Explanation:
The given parameters, are;
The location of q₁ = The origin
The location of q₂ = 14.7 mm from q₁
The repulsive force exerted on q₂ by q₁ = 2.62 μN
The location the particle q₂ is located to = 18.0 mm from q₁
By Coulomb's law, we have;
Where;
k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)
r = The distance between the particles
F = The force acting between the particles
When r = 14.7 mm F = 2.62 μN
∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸
q₂ × q₁ = 1.48 × 10⁻¹⁸ C²
When the distance is increased to 18.0 mm, we have;
F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N
∴ F = 1.75 μN.
Therefore;
The force magnitude is 1.75 μN outward from the origin towards x₂.