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Stolb23 [73]
3 years ago
10

You want to manufacture a guitar such that the instrument will be in tune when each of the strings are tightened to the same ten

sion. The middle (D) string on the guitar should have fundamental frequency 146.83 Hz. The highest (E) string should have fundamental frequency 329.63 Hz. If the D string has linear mass density 0.00256kg/m, what should be the mass density of the E string
Physics
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

0.000507 kg/m

Explanation:

L = Length of string

T = Tension

\mu = Mass density of string

E denotes the E string

D denotes the D String

Frequency is given by

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

So

f\propto \sqrt{\dfrac{1}{\mu}}

\dfrac{f_D}{f_E}=\sqrt{\dfrac{\mu_E}{\mu_D}}\\\Rightarrow \mu_E=\dfrac{f_D^2}{f_E^2}\mu_D\\\Rightarrow \mu_E=\dfrac{146.83^2}{329.63^2}\times 0.00256\\\Rightarrow \mu_E=0.000507\ kg/m

The mass density of the E string is 0.000507 kg/m

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Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

4 0
3 years ago
(URGENT- Due today.)
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Answer:

I=\sqrt{} \frac{P}{R}

Explanation:

6 0
3 years ago
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0
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Your answer would be:

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3 years ago
uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0
olganol [36]

Answer:

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This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0
3 years ago
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