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)
Answer:
From smallest ratio to the largest ratio:
Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)
Explanation:
The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1
The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1
Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1
Answer:
The answer is A
Explanation:
The octopus’s tentacle keeps moving right after it is bitten off
Answer:
1. 310 N
2. 310 N
Explanation:
1. Determination of the net force.
Force applied (Fₐ) = 430 N
Force experience (Fₑ) = 120 N
Net force (Fₙ) =?
Fₙ = Fₐ – Fₑ
Fₙ = 430 – 120
Fₙ = 310 N
Hence, the net force acting on the crate is 310 N
2. Determination of the net force.
Force applied (Fₐ) = 375 N
Force of friction (Fբ) = 65 N
Net force (Fₙ) =?
Fₙ = Fₐ – Fբ
Fₙ = 375 – 65
Fₙ = 310 N
Hence, the net force acting on the crate is 310 N.
