Answer:0.061
Explanation:
Given
Temperature of soup 
heat capacity of soup 
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by
integrating From 
to 
![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by
Fraction of the total heat that is lost by the soup can be turned is given by
![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)
The picture is kinda blurry but it looks like granite which is metamorphic.
Answer:
B) 4500 Pa
Explanation:
As pressure is force per unit area,
P = F/A
It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.
The possible areas are
0.30(0.40) = 0.12 m²
0.30(0.50) = 0.15 m²
0.40(0.50) = 0.20 m²
The pressure when the face with the largest area (0.20 m²) is down is
P = 900 / 0.20 = 4500 N/m² or 4500 Pa
the other possible pressures would be
900/0.15 = 6000 Pa
900/0.12 = 7500 Pa
which are both larger than our solution.
D. email is saved on the server that transfers is
