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elixir [45]
2 years ago
13

A 60 year old person has a threshold of hearing of 79.0 dB for a sound with frequency f=10,000 Hz. By what factor must the inten

sity of a sound wave of that frequency, audible to a typical young adult, (sound level=43.0 dB) be increased so that it is heard by the older person.
Physics
1 answer:
liubo4ka [24]2 years ago
6 0

They provided the intensity in decibels for the problem, but they are unsure by what factor to increase it (I) to make the sound loud enough for the elderly person to hear.

Neglect f entirely.

The following equation must be used to convert decibels (dB) to I:

I=(10^(dB/10))*10^-12

Divide the elder person's dB by the younger person's dB after doing this for each dB.

After putting values and solving

we get,

1.8372093023.

So, the final answer is 1.8372093023.

learn more about frequency brainly.com/question/254161

#SPJ1

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A sound wave travels twice as far in neon (Ne) as it does it krypton (Kr) in the same time interval. Both neon and krypton can b
avanturin [10]

Answer:

281.6 K

Explanation:

The speed of sound in an ideal gas is given by c = √(γKT/m).

From the question speed of sound in Ne, c₁ = 2c₂ speed of sound in Kr

c₁ = √(γKT₁/m₁) and c₂ = √(γKT₂/m₂)

So  √(γKT₁/m₁)  = 2√(γKT₂/m₂) where T₁, m₁ and T₂, m₂ are the temperatures and atomic masses of Neon and Krypton respectively.

So, √(T₁/m₁)  = 2√(T₂/m₂)

(T₁/m₁)  = 4(T₂/m₂) (squaring both sides)

T₁ = 4(T₂m₁/m₂)

Given that m₁ = 20.2 u , m₂ = 83.8 u, T₂ = 292 K

T₁ = 4(292 × 20.2/83.8) K = 23593.6/83.8 = 281.55 K ≅ 281.6 K

6 0
3 years ago
A plastic light pipe has an index of refraction of 1.48. For total internal reflection, what is the minimum angle of incidence i
Rama09 [41]

Answer:

a) 42.52°

b) 63.98°

Explanation:

Refractive index of pipe = 1.48 = n₂

Refractive index of air = 1.0003 = n₁

\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{1.0003}{1.48}\\\Rightarrow \theta_r=sin^{-1}0.676\\\Rightarrow \theta_r=42.52^{\circ}

∴ Minimum angle of incidence is 42.52°

Refractive index of water = 1.33 = n₁

\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{1.33}{1.48}\\\Rightarrow \theta_r=sin^{-1}0.899\\\Rightarrow \theta_r=63.98^{\circ}

∴ Minimum angle of incidence is 63.98°

5 0
4 years ago
Electrical current is defined as
lbvjy [14]

The answer is B. It is measured in amperes.

5 0
3 years ago
13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra
Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

\frac{dv}{dt} = a.

solving the last equation

\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at,

where v_0=0 because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt

then

x- x_0 = \frac{1}{2}at^2

where x_0=0 if we consider  the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If x=1.11\ m at t=1s, then

a = \frac{2x}{t^2} = 2.22\ m/s^2

and the final expression for the distance is

x = 1.11 t^2.

If t = 2s, x = 4.44 m. Which means thad the additional distance is

x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m

8 0
4 years ago
If a net force of 800n is acting on an object with a mass of 60kg. What would be the objects accelerations
aivan3 [116]

Answer: 13.33 m/s^2

Explanation:

Force is the product of mass of an object and its acceleration.

i.e Force = Mass x Acceleration

Since Force on object = 800N

Mass of object = 6Kg

Acceleration = ?

Then, Force = Mass x Acceleration

800N = 60Kg x Acceleration

Acceleration = (800N/60kg)

Acceleration = 13.33 m/s^2 (The unit of acceleration is metre per second square)

Thus, the objects accelerations is 13.33 m/s^2

5 0
4 years ago
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