I’m not sure sorry I really wish I could help
Answer:
The length is 
Explanation:
From the question we are told that
The length of the hand is 
The angle at the hand is held is
Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

substituting values


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)

Your answer would be : Leyden jar.