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3 years ago

15

A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

ng sensitivity?

1 answer:

ElenaW [278]3 years ago

7 0

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency c/λ where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

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3 years ago

What are you good at and hobbies?

alisha [4.7K]

Answer: well im a soccer player, im a competitive cheerleader and I love reading and writing

Explanation:

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3 years ago

Catalytic ozone destruction occurs in the stratosphere by a two-step reaction:

Sladkaya [172]

Catalytic ozone destruction occurs in the stratosphere where the reactions involving bromine, chlorine, hydrogen, nitrogen and oxygen gases form compounds that destroy the ozone layer. The reactions uses a catalyst (speeds up the reaction) in a two step reaction. considering chlorine the reactions appears as follows;
step 1
Cl + O3 = ClO + O2
step 2
ClO + O = Cl + O2
Where by chlorine is released to destroy the ozone layer, this takes place many times even with the other elements (hydrogen, bromine, nitrogen) and the end result is a completely destroyed Ozone layer

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3 years ago

Please help don't for get to show work

emmasim [6.3K]

-- Bob covered a distance of (32m + 45m) = 77 meters.

-- His displacement is the straight-line distance and direction
from his starting point to his ending point.

The straight-line distance is

D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters

The direction is the angle whose tangent is (32/45) south of east.

tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.

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3 years ago

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

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3 years ago