Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³
Therefore:
8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.
Answer: 8.06 in³=1.321 x 10⁻⁴ m³
The speed of the block when the compression is 15 cm is 9.85 m/s.
The given parameters;
- <em>mass of the block, m = 2.4 kg</em>
- <em>height of the block, h = 5 m</em>
- <em>compression of the spring, x = 25 cm = 0.25 m</em>
The spring constant is calculated as follows;

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.
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I'm pretty sure it's B
I hope that helps
Answer:
Therefore,
The potential (in V) near its surface is 186.13 Volt.
Explanation:
Given:
Diameter of sphere,
d= 0.29 cm


Charge ,

To Find:
Electric potential , V = ?
Solution:
Electric Potential at point surface is given as,

Where,
V= Electric potential,
ε0 = permeability free space = 8.85 × 10–12 F/m
Q = Charge
r = Radius
Substituting the values we get


Therefore,
The potential (in V) near its surface is 186.13 Volt.