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disa [49]
3 years ago
14

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

drums, which lie on the surface of the head. And, surprisingly, the sound has nothing to do with the frog's inflated throat. If the emitted sound has a frequency of 262 Hz and a sound level of 84 dB (near the eardrum), what is the amplitude of the eardrum's oscillation? The air density is 1.21 kg/m3 and the speed of sound is 346 m/s.
Physics
1 answer:
Sidana [21]3 years ago
8 0

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

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The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie
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Complete Question

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Answer:

The uncertainty in inverse frequency is  \Delta  [\frac{1}{w} ]=  \frac{3}{2000} \ s

Explanation:

From the question we are told that

   The value of the proportionality constant is  k  = 5  \frac{Hz }{T}

   The strength of the magnetic field is  B = 20 \ T

   The change in this strength of magnetic field is  \Delta B = 3  \ T

The magnetic field is given as

           B  =  \frac{k}{\frac{1}{w} }

Where w is frequency

The uncertainty or error of the field is given as

         \Delta  B  =  \frac{k }{[\frac{1}{w}^]^2 }  \Delta [\frac{1}{w} ]

The uncertainty in inverse frequency is given  as

           \Delta  [\frac{1}{w} ]  = \frac{\Delta B}{k [\frac{1}{w^2} ]}

                    \Delta  [\frac{1}{w} ]=  \frac{\Delta B}{k (B)^2 }

substituting values

                  \Delta  [\frac{1}{w} ]=  \frac{3}{5 (20)^2 }

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6 0
3 years ago
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
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Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

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Boltzmann’s constant is 1.38066×10−23 J/K, and the universal gas constant is 8.31451 J/K · mol. If 1.9 mol of a gas is confined
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Answer:

The average kinetic energy of a gas molecule is 8.49 × 10⁻²¹ J

Explanation:

According to Kinetic Molecular Theory, the average kinetic energy of gas molecules is a function only of temperature and is given by the equation:

E=\frac{3}{2}kT

pV=nRT;\\T=\frac{pV}{nR}

Substituting the value of T in the kinetic energy equation:

E=\frac{3kpV}{2nR}

Where p is the pressure of the gas = 9 atm = 9 × 101325 pa

k is the Boltzmann’s constant  =  1.38066 × 10⁻²³ J/K,

V is the volume of the gas = 7.1 L = 7.1 × 10⁻³ m³

n is the number of moles of the gas = 1.9 mol

R is the universal gas constant = 8.31451 J/K · mol.

E is the average kinetic energy of a gas molecule

Substituting values:

E=\frac{3kpV}{2nR}=\frac{3*1.38066*10^{-23}*9*101325*7.1*10^{-3}}{2*1.9*8.31451}=8.49*10^{-21}J

The average kinetic energy of a gas molecule is 8.49 × 10⁻²¹ J

6 0
3 years ago
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