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3 years ago

By what factor does the sound intensity increase if the sound intensity level increases from 60 db to 61 db?

Physics

1 answer:

tiny-mole [99]3 years ago

6 0

Answer:

$\frac{I_{2}}{I_{1}} = 10^{0.1}$

Explanation:

The sound intensity level β can be mathematically defined as

$\beta(dB) = 10log_{10}(\frac{I}{I_{o} } )$

where $I_{o}$ is the reference intensity and is equal to $10^{-12}$ $W/m^{2}$ .

Now,

$\beta_{2} - \beta_{1} = 10 log\frac{I_{2} }{I_{o}} - 10 log\frac{I_{1} }{I_{o}}\\62-60= 10log(\frac{\frac{I_{2} }{I_{o} } }{\frac{I_{1} }{I_{o} } } )\\1=10log(\frac{I_{2} }{I_{1} } )\\\\\frac{I_{2} }{I_{1} }=10^{0.1}$

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Answer:

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the correct answer is option E

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3 years ago

2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela

Vanyuwa [196]

Answer:

A) E = 145.6 N / C , B) y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D) F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

F = m a

- e E = m a

a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

x = v₀ t

t = x / v₀

y axis (perpendicular to plates)

y = y₀ + $v_{oy}$ t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

y = ½ a t²

we substitute

y = ½ (e E /m) (x / v₀)²

y = ½ e x2 /m v₀² E

we look for the electric field

E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

E = 2 9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

y = ½ e x² / m v₀² E

y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)² 145.6

y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

$F_{g}$ = G m M / R²