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kogti [31]
4 years ago
5

A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a

pproaches and b) recedes from the musician with a speed of 11.0 m/s?
Physics
1 answer:
DanielleElmas [232]4 years ago
6 0

Answer:

a) f_o=454.11Hz

b)f_o=425.89Hz

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

f_o=\frac{v\pm v_o}{v\pm v_s} *f_s

where:

f_o=Observed\hspace{3}frequency

f_s=Actual\hspace{3}frequency

v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves

v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source

v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer

Now let's consider the next cases:

+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source

-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source

-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer

+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer

The data provided by the problem is:

f_s=440Hz\\v_o=11m/s

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

v=343m/s

Now, in the first case the observer alone is in motion towards to the source, hence:

f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz

Finally, in the second case the observer alone is in motion away from the source, so:

f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz

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gtnhenbr [62]

Answer:

V=2.4\times 10^2\ \text{m}^3

Explanation:

Given that,

The dimensions of a room are 16.40 m long, 4.5 m wide and 3.26 m high.

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The room in the shape of cuboid whose volume is given by :

V=L\times B\times H

L,B,H are length, breadth and height of the room

So,

V=16.4\times 4.5\times 3.26\\\\V=240.588\ \text{m}^3\\\\\text{In scientific notation}\\\\V=2.4\times 10^2\ \text{m}^3

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5 0
3 years ago
If the tensile strength of the Kevlar 49 fibers is 0.550 x 106 psi and that of epoxy resin is 11.0 x103 psi, calculate the stren
Elina [12.6K]

Answer:

\sigma_{tot} =436810 psi

Explanation:

Notation

\sigma_{fib} =0.55x10^6 psi represent the tensile strength for the Kevlar 49 fibers

\sigma_{epox} =11x10^3 psi represent the tensile strength for the epoxy resin

79% by volume is made of Kevlar 49 fibers

The rest 100-79=21% is made of epoxy resin

Tensile strength is a "measurement of the force required to pull something such as rope, wire, or a structural beam to the point where it breaks". Can be defined as "the maximum amount of tensile stress that it can take before failure".

For this case we need to use a basis of calculus since we don't know the total volume but we can assume a reference value on order to make the calculations.

If we assume a total volume of V_{tot}=1 in^3 we can do the follwoing balance:

\sigma_{tot}V_{tot}=\sigma_{fib}V_{fib} +\sigma_{epox}V_{epox}

We can replace the values given:

\sigma_{tot}=(0.55x10^6 psi)(0.79) +(11x10^3 psi)(1-0.79)

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4 years ago
A 2.40 kg can of coffee moving at 1.50 m/s in the +x-direction on a kitchen counter collides head-on with a 1.20 kg box of macar
liraira [26]

Answer:

The macaroni travels with a velocity of 1.35 m/s in the +x-direction.

Explanation:

By the conservation of momentum principle and assuming there is no external force (e.g. friction) acting on the system, the total initial momentum is equal to the total final momentum.

We take all velocities in the +x-direction as positive.

Initial momentum of coffee can = (2.40 kg)(1.50 m/s) = 3.60 kg m/s

Initial momentum of macaroni = (1.20 kg)(0.00 m/s) = 0.00 kg m/s

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Since this is positive, the macaroni travels with a velocity of 1.35 m/s in the +x-direction.

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