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Leto
7 days ago
14

The G string on a guitar is 69 cm long and has a fundamental frequency of 196 Hz. A guitarist can play different notes by pushin

g the string against various frets, which changes the string's length. The third fret from the neck gives B♭ (233.08 Hz); the fourth fret gives B (246.94 Hz).
How far apart are the third and fourth frets?

Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Keith_Richards [2.9K]7 days ago
6 0
Δd = 23 cm. When the eta string of the guitar has nodes at both ends, the resulting waves create a standing wave, which can be expressed with the following formulas: Fundamental: L = ½ λ, 1st harmonic: L = 2 ( λ / 2), 2nd harmonic: L = 3 ( λ / 2), Harmonic n: L = n λ / 2, where n is an integer. The rope's speed can be calculated using the formula v = λ f. This speed remains constant based on the tension and linear density of the rope. Now, let's determine the speed with the provided data: v = 0.69 × 196, yielding v = 135.24 m/s. Next, we will find the wavelengths for the two frequencies: λ₁ = v / f₁, which gives λ₁ = 135.24 / 233.08, equaling λ₁ = 0.58022 m; λ₂ = v / f₂ results in λ₂ = 135.24 / 246.94, consequently λ₂ = 0.54766 m. We'll substitute into the resonance equation Lₙ = n λ/2. At the third fret, m = 3, therefore L₃ = 3 × 0.58022 / 2, resulting in L₃ = 0.87033 m. For the fourth fret, m = 4, which gives L₄ = 4 × 0.54766 / 2, equating to L₄ = 1.09532 m. The distance between the two frets is Δd = L₄ – L₃, so Δd = 1.09532 - 0.87033, leading to Δd = 0.22499 m or 22.5 cm, rounded to 23 cm.
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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [2668]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
1 month ago
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
Softa [2668]
The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s

Conclusion: 12.3 m/s
7 0
1 month ago
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In an inertial frame of reference, a series of experiments is conducted. in each experiment, two or three forces are applied to
Yuliya22 [2968]
Objects will stay in a stationary position if the total force acting on them amounts to zero; this occurs when equal forces are applied in opposite directions. According to Newton's second law, if the net force on an object is zero, it will not move.
8 0
29 days ago
A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
ValentinkaMS [3091]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

  • 52.92% metal: 52.92 g M
  • 47.80% oxygen: 47.80 g O

 By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

moles of oxygen=\frac{47.8g*1mol}{16g}

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

Given 52.92 g of metal in the sample, the molar mass of the metal is:

molar mass=\frac{52.92 g}{5.975 mol}

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0
16 days ago
A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/
ValentinkaMS [3091]

The formula for range is:

R = \frac{v_o^2 sin2\theta}{g}

Given values are:

v_0=37.2m/s

where θ equals 14.1 degrees

g=9.80m/s^2

Using the equation above,

R = \frac{37.2^2 sin2*14.1}{9.80}

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0
1 month ago
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