Abdou was explaining to a classmate that graphite is a good lubricant
1 answer:
The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other.
<h3>Why is graphite a solid lubricant?</h3>We know that graphite is composed of layers. These hexagonal layers are held together by weak Van Der Walls forces and as such are able to slide over each other. The carbon atom in each layer are held together by strong covalent bonds.
The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other and as such make the graphite fluid.
Thus, the image that shows these layers of graphite is attached to this an answer
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Answer:
<u>Part A</u>
I = 1.4 mW/m²
<u>Part B</u>
β = 91.46 dB
Explanation:
<u>Part A</u>
Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.
For a spherical sound wave, the sound intensity is given by;
Where;
P is the source of power in watts (W)
I is the intensity of the sound in watt per square meter (W/m2)
r is the distance r away
Given:
P = 34 W,
A = 1.0 cm²
r = 44 m
The sound intensity at the position of the microphone is calculated to be;
I = 0.0013975 W/m²
≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²
I = 1.4 mW/m²
The sound intensity at the position of the microphone is 1.4 mW/m².
<u>Part B</u>
Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.
Sound intensity level is calculated as;
β dB
Where,
β is the Sound intensity level in decibels (dB)
I is the sound intensity;
I₀ is the reference sound intensity;
By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²
∴ β
β
β = 91.46 dB
The sound intensity level at the position of the microphone is 91.46 dB.