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balandron [24]
4 years ago
8

The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu

ndamental frequency around which we would expect hearing to be most sensitive? Assume the speed of sound in air to be 335 m/s.
Physics
1 answer:
zhenek [66]4 years ago
3 0

Answer:

f=2887.93Hz

Explanation:

Given data

Length L=2.9 cm=0.029m

Speed of sound v=335 m/s

to find

Fundamental frequency f

Solution

As we know that frequency is given as:

f=v/λ

f=\frac{v}{4L}\\ f=\frac{335m/s}{4(0.029m)} \\ f=2887.93Hz

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The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3
bagirrra123 [75]

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

8 0
3 years ago
A refrigerator removes 55.0 kcal of heat from the freezer and releases 73.5 kcal through the condenser on the back.How much work
sammy [17]

Here refrigerator removes 55 kcal heat from freezer

Refrigerator releases 73.5 kcal heat to surrounding

So here we can use energy conservation principle by II Law of thermodynamics

the law says that

Q_1 = Q_2 + W

here we know that

Q_1 = heat released to the surrounding

Q_2 = heat absorbed from freezer

W = work done by the compressor

now using above equation we can write

73.5 = 55 + W

W = 73.5 - 55

W = 18.5 kcal

So here compressor has to do 18.5 k cal work on it

5 0
3 years ago
4. While broadcasting a football game, the announcer exclaimed, "I can't believe it. Carl James just scored a touchdown. That's
Zielflug [23.3K]
D. Carl doesnt score touchdowns very often.
7 0
4 years ago
Read 2 more answers
a boy is riding a bicycle at a velocity of 5.0 m/s. he applies the brakes and uniformly decelerates to a stop at a rate of 2.5 m
JulijaS [17]

The working equation would be Vf (final velocity) = Vi (initial velocity) + a (acceleration) t (time). The given data are the initial velocity (5.0 m/s), acceleration (-2.5 m/s^2, negative since it is said to decelerate) and the final velocity (0 m/s, since it will put to a stop). The time would be 2 seconds. 

3 0
4 years ago
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