Six centimeters equal to about two inches
Answer:
The distance is 
Explanation:
From the question we are told that
The distance from the conversation is 
The intensity of the sound at your position is 
The intensity at the sound at the new position is 
Generally the intensity in decibel is is mathematically represented as
![\beta = 10dB log_{10}[\frac{d}{d_o} ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%2010dB%20log_%7B10%7D%5B%5Cfrac%7Bd%7D%7Bd_o%7D%20%5D)
The intensity is also mathematically represented as

So
![\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%2010dB%20%2A%20%20log_%7B10%7D%5B%5Cfrac%7BP%7D%7BA%2A%20d_o%7D%20%5D)
=> ![\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cbeta%7D%7B10%7D%20%20%3D%20%20log_%7B10%7D%20%5B%5Cfrac%7BP%7D%7BA%20%28l_o%29%7D%20%5D)
From the logarithm definition
=> 
=> ![P = A (d_o ) [10^{\frac{\beta }{ 10} } ]](https://tex.z-dn.net/?f=P%20%3D%20%20A%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta%20%7D%7B%2010%7D%20%7D%20%5D)
Here P is the power of the sound wave
and A is the cross-sectional area of the sound wave which is generally in spherical form
Now the power of the sound wave at the first position is mathematically represented as
![P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]](https://tex.z-dn.net/?f=P_1%20%3D%20%20A_1%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_1%20%7D%7B%2010%7D%20%7D%20%5D)
Now the power of the sound wave at the second position is mathematically represented as
![P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20A_2%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_2%20%7D%7B%2010%7D%20%7D%20%5D)
Generally power of the wave is constant at both positions so
![A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]](https://tex.z-dn.net/?f=A_1%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_1%20%7D%7B%2010%7D%20%7D%20%5D%20%20%3D%20A_2%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_2%20%7D%7B%2010%7D%20%7D%20%5D)
![4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]](https://tex.z-dn.net/?f=4%20%5Cpi%20r_1%20%5E2%20%20%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_1%20%7D%7B%2010%7D%20%7D%20%5D%20%20%3D%204%20%5Cpi%20r_2%20%5E2%20%20%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_2%20%7D%7B%2010%7D%20%7D%20%5D)
![r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}](https://tex.z-dn.net/?f=r_2%20%3D%20%20%5Csqrt%7Br_1%20%5E2%20%5B%5Cfrac%7B10%5E%7B%5Cfrac%7B%5Cbeta_1%7D%7B10%7D%20%7D%7D%7B%2010%5E%7B%5Cfrac%7B%5Cbeta_2%7D%7B10%7D%20%7D%7D%20%5D%7D)
substituting value
![r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}](https://tex.z-dn.net/?f=r_2%20%3D%20%20%20%5Csqrt%7B%2024%5E2%20%5B%5Cfrac%7B10%5E%7B%5Cfrac%7B%2040%7D%7B10%7D%20%7D%7D%7B10%5E%7B%5Cfrac%7B80%7D%7B10%7D%20%7D%7D%20%5D%7D)

Answer:
True plz thank me this is the answer
The weight of the elevator is
W = (1500 kg)*(9.8 m/s²) = 1.47 x 10⁴ N
Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is
T = W + R = 1.47 x 10⁴ + 100 = 1.48 x 10⁴ N
By definition, the work done in lifting the elevator by 40 m is
(1.48 x 10⁴ N)*(40 m) = 5.92 x 10⁵ J = 592 kJ
Answer: 592 kJ (or 5.92 x 10⁵ J)
V = speed of the plane with no wind = 250 mph
v = speed of wind
consider the motion of the plane while traveling with the wind :
V' = speed of plane travelling in the direction of wind = V + v = 250 + v
d' = distance traveled = 300 miles
t' = time taken to travel the distance
time taken to travel the distance is given as
t' = d'/V'
t' = 300/(250 + v) eq-1
consider the motion of the plane while traveling against the wind :
V'' = speed of plane travelling against the direction of wind = V - v = 250 - v
d'' = distance traveled = 200 miles
t'' = time taken to travel the distance
time taken to travel the distance is given as
t'' = d''/V''
t'' = 200/(250 - v) eq-2
Given that :
time taken to travel 300 miles = time taken to travel 200 miles
t' = t''
using eq-1 and eq-2
300/(250 + v) = 200/(250 - v)
3/(250 + v) = 2/(250 - v)
750 - 3 v = 500 + 2 v
5 v = 250
v = 50 mph