Hello!
The winds affected by specific landforms on earth's surface are: Local winds.
I hope my answer helped you out! :)
The answer is B artificial selection
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
So the ratio will be ![\frac{T_L}{T_H}=-0.171](https://tex.z-dn.net/?f=%5Cfrac%7BT_L%7D%7BT_H%7D%3D-0.171)
Explanation:
We have given heat engine absorbs 450 joule from high temperature reservoir
So ![Q=450j](https://tex.z-dn.net/?f=Q%3D450j)
As the heat engine expels 290 j
So work done W = 290 J
We know that efficiency ![\eta =\frac{W}{Q}=\frac{290}{450}=0.6444](https://tex.z-dn.net/?f=%5Ceta%20%3D%5Cfrac%7BW%7D%7BQ%7D%3D%5Cfrac%7B290%7D%7B450%7D%3D0.6444)
It is given that efficiency of the engine only 55 % of Carnot engine
So efficiency of Carnot engine ![=\frac{0.6444}{0.55}=1.171](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.6444%7D%7B0.55%7D%3D1.171)
Efficiency of Carnot engine is ![\eta =1-\frac{T_L}{T_H}](https://tex.z-dn.net/?f=%5Ceta%20%3D1-%5Cfrac%7BT_L%7D%7BT_H%7D)
![1.171 =1-\frac{T_L}{T_H}](https://tex.z-dn.net/?f=1.171%20%3D1-%5Cfrac%7BT_L%7D%7BT_H%7D)
![\frac{T_L}{T_H}=-0.171](https://tex.z-dn.net/?f=%5Cfrac%7BT_L%7D%7BT_H%7D%3D-0.171)