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Darina [25.2K]
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?
Physics
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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Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

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At which angle must a laser beam enter the water for no refraction to occur?
abruzzese [7]

Light that enters the new medium <em>perpendicular to the surface</em> keeps sailing straight through the new medium unrefracted (in the same direction).

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Moral of the story:  If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.

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If you wanted to produce more power,what could you do to maximize power
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6 0
3 years ago
Read 2 more answers
Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o
KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹  N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴  +  -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0
3 years ago
A grating has 2000 slits/cm/cm. How many full spectral orders can be seen (400 to 700 nmnm) when it is illuminated by white ligh
Anton [14]

With the use of the formula SinФ = nλ / d, there are 16 spectral orders which can be seen when it is illuminated by white light.

Given that a grating has 2000 slits/cm. That is,

d = 0.01 / 2000

d = 5 x 10^{-6} m

The wavelength λ = (700 - 400) nm

λ = 300 x 10^{-9} m

To calculate how many full spectral orders that can be seen (400 to 700 nm) when it is illuminated by white light, we will use the below formula

SinФ = nλ / d

Φ = Sin^{-1} (nλ / d)

When n = 1

Φ = Sin^{-1}(300 x 10^{-9}/ 5 x 10^{-6} )

Φ = 3.4 degrees

when n = 2

Φ = Sin^{-1}(2 x 300 x 10^{-9}/ 5 x 10^{-6} )

Ф = 6.9 degrees

When n = 3

Ф = Sin^{-1}(3 x 300 x 10^{-9}/ 5 x 10^{-6} )

When n = 16

Ф = Sin^{-1}(16 x 300 x 10^{-9}/ 5 x 10^{-6} )

Ф = Sin^{-1}(0.96)

Ф = 73.7 degrees

when n = 17

Ф = Sin^{-1}(17 x 300 x 10^{-9}/ 5 x 10^{-6} )

Ф = Sin^{-1}(1.05)

Ф = Error ( that is, it does not exist)

Therefore, there are 16 spectral orders which can be seen when it is illuminated by white light.

Learn more about double slit here: brainly.com/question/4449144

6 0
2 years ago
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