Answer:
Point 3
Explanation:
It is a summer month position but it has no sunlight.
Answer:
The value of change in internal energy of the gas = + 1850 J
Explanation:
Work done on the gas (W) = - 1850 J
Negative sign is due to work done on the system.
From the first law we know that Q = Δ U + W ------------- (1)
Where Q = Heat transfer to the gas
Δ U = Change in internal energy of the gas
W = work done on the gas
Since it is adiabatic compression of the gas so heat transfer to the gas is zero.
⇒ Q = 0
So from equation (1)
⇒ Δ U = - W ----------------- (2)
⇒ W = - 1850 J (Given)
⇒ Δ U = - (- 1850)
⇒ Δ U = + 1850 J
This is the value of change in internal energy of the gas.
Answer:
B) boiling point
Explanation:
The movement of the particles causes the shape of the liquid to change. The liquid will flow and fill to the lowest part of the container, in the shape of the container
But the volume does not change. The limited amount of space between the particles means that the liquid has only very limited compressibility.
iron, cobalt and Nickel is the answer
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)
