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3 years ago

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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.

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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Explanation:

For this exercise let's use the constructor equation

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In a mirror the focal length is

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indicate us radius of curvature is equal to the diameter of the eye

R = 3,50 10² mm

f = 3.50 10² /2 = 1.75 10² mm

they also say that the distance to the object is p = 0.800 10³ mm

1 / q = 1 / f - 1 / p

1 / q = 1 / 175 - 1 /800

1 / q = 0.004464

q = 224 mm

to calculate the size let's use the magnification ratio

m = $\frac{h'}{h} = - \frac{q}{p}$

h '= $- \frac{q}{p} \ h$

h ’= - 224 350 / 800

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in concave mirrors the image is real.

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3 years ago

Explain Newton's third law in your own words.

Kazeer [188]

Newton's third law states "for every action, there is an equal and opposite reaction."

What this is pretty much saying is that for every action, there is a consequence. One force connects and triggers another.

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3 years ago

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

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Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

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while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

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3 years ago

Part of one row of the periodic table is shown. 4 blue boxes in a row. The first has S n at the center and 50 above with tin and

nydimaria [60]

Answer:

Atoms of tellurium (Te) have the greatest average number of neutrons equal to 76.

Explanation:

In the periodic table, Elements are represented with their respected symbols. Above the symbol is the elements atomic number which is equal to the number of protons in each atom. Below the symbol is the mass number of that element which is roughly equal to the sum of neutrons and protons of that atom.

To calculate the number of neutrons we can take the difference of Atomic number and mass number:

Number of neutrons = mass number - atomic number

<u>- Tin:</u>

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Number of neutrons = mass number - atomic number = 127 - 53

Number of neutrons = 74

Here, the greatest number of neutrons is for the atoms of Tellurium(Te).

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3 years ago

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frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

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$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago