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aev [14]
3 years ago
9

A composite material is to be designed with epoxy (Em 3.5 GPa) and unidirectional fibers. The longitudinal elastic modulus of th

e designed composite material is to be at least E 320 GPa. If the fiber volume fraction is 70%, calculate the minimum elastic modulus requirement of the fiber material.
Physics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

Minimum elastic modulus of fiber = 455.64 GPa

Explanation:

Contents of composite material = Epoxy and Unidirectional fibers

Elastic modulus of epoxy = 3.5 GPa

Elastic modulus of composite material = 320 GPa

Volume fraction of fiber = 70 %

Volume fraction of epoxy = 100 - 70 = 30%

Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320

0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95

Elastic modulus of fiber = 455.64 GPa

Minimum elastic modulus of fiber = 455.64 GPa

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D - tertiary consumer

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A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the
olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

f = frequency of the tone = 329.6 Hz

T = Time period of the sound wave

we know that, Time period and frequency are related as

T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s

v = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m

v = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m

8 0
3 years ago
Plz hurry How fast can work be done by a 7 horsepower engine?
den301095 [7]

Answer:

It will go about 45-50 mph.

Explanation:

Depending on what you are hauling.

7 0
3 years ago
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What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
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Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

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A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

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Mass, M = \rm 2.1\times 10^{-22}\ g.

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Therefore, the density of the 127 I nucleus is given by

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On comparing with the density of the solid iodine,

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7 0
3 years ago
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