Answer:
The value is 
Explanation:
From the question we are told that
The efficiency of the carnot engine is 
The efficiency of a heat engine is 
The operating temperatures of the carnot engine is
to 
The rate at which the heat engine absorbs energy is 
Generally the efficiency of the carnot engine is mathematically represented as
![\eta = [ 1 - \frac{T_1 }{T_2} ]](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%20%5B%201%20-%20%5Cfrac%7BT_1%20%7D%7BT_2%7D%20%20%5D)
=> ![\eta = [ \frac{T_2 - T_1}{T_2} ]](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%20%5B%20%5Cfrac%7BT_2%20-%20T_1%7D%7BT_2%7D%20%5D)
=> 
Generally the efficiency of the heat engine is

=> 
Generally the efficiency of the heat engine is also mathematically represented as

Here W is the work done which is mathematically represented as

Here
is the heat exhausted
So

=> 
=> 
Answer:
OHHH.. I know what you talking about.
Explanation:
Okay so Netforce escape room is a Game on the playstore app and every time you type the code in it unlocks the door and you move up to different levels.
Achieve a full outer shell
Answer:

Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by

Substitute the values then we get
Capacitance between parallel plate capacitor=

V=


Hence, the charge resides on the outer surface=