You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?

  r₂ = 1,586 m

Explanation:

For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m

         β = 10 log (I / I₀)

where Io is the sensitivity threshold 10⁻¹² W / m²

          I₁ / I₀ =

          I₁ = I₀  e^{\beta/10}

let's calculate

          I₁ = 10⁻¹² e^{25/10}

          I₁ = 1.20 10⁻¹¹ W / m²

the other intensity in exercise is

          I₂ = 10⁻¹² e^{80/10}

          I₂ = 2.98 10⁻⁹ W / m²

now we use the definition of sound intensity

          I = P / A

where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed

         P = I A

the area a sphere is

         A = 4π r²

we can write this equation for two points of the found intensities

          I₁ A₁ = I₂ A₂

where index 1 corresponds to 25m and index 2 to the other distance

          I₁ 4π r₁² = I₂ 4π r₂²

          I₁ r₁² = I₂ r₂²

           r₂ = √ (I₁ / I₂) r₁

let's calculate

           r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25

           r₂ = √ (0.40268 10⁻²) 25

           r₂ = 1,586 m

tbh i dont know ether

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s