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m_a_m_a [10]
3 years ago
5

What is the pressure amplitude for the threshold of human hearing?

Physics
2 answers:
Hoochie [10]3 years ago
5 0
The pressure amplitude for the threshold of human hearing is"<span> RMS sound pressure of </span>20 micropascals, corresponding to a sound intensity of 0.98 pW/m2<span> at </span>1 atmosphere<span> and 25 °C"</span>
Korolek [52]3 years ago
4 0

The answer is B,

1.0 x 10^-12 watts/meter^2

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/intens.html

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Which of these stars has the hottest core?A)a blue main-sequnce star. B)a red supergiant. C)a red main-sequence star.
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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t
Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                                P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                               P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                               P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                      = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                  ∴                 K.E = P.E

                                     1/2 mv² = 138.44

                                     1/2 x 25 x v² 138.44

                                            v² = 11.0752

                                             v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

             Tension on string, T = Force acting on the swing, F

                      W=L\int\limits^0_\phi{F} \, d \phi

                             =L\int\limits^0_\phi{mg.sin \phi} \, d \phi

                            = -Lmg[cos\phi]_{42}^{0}

                            = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                            = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0
3 years ago
In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0
3 years ago
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