Answer:
The engine consists of a fixed cylinder and a moving piston. The expanding combustion gases push the piston, which in turn rotates the crankshaft. Ultimately, through a system of gears in the power-train, this motion drives the vehicle's wheels.
Explanation:
Answer:
month = input("Input the month (e.g. January, February etc.): ")
day = int(input("Input the day: "))
if month in ('January', 'February', 'March'):
season = 'winter'
elif month in ('April', 'May', 'June'):
season = 'spring'
elif month in ('July', 'August', 'September'):
season = 'summer'
else:
season = 'autumn'
if (month == 'March') and (day > 19):
season = 'spring'
elif (month == 'June') and (day > 20):
season = 'summer'
elif (month == 'September') and (day > 21):
season = 'autumn'
elif (month == 'December') and (day > 20):
season = 'winter'
print("Season is",season)
Explanation:
Answer: True
Explanation: Closed loop relies on feedback from PNS to make modifications in the movement, open loop allows action in the absence of feedback, 2. ... Closed loop can change the initial commands, open loop can not change the initial commands.
Answer:
Distribution factor P = =38.33
V = 7.826 ml
Explanation:
given details:
BOD =230 mg/l
DO inital = 8.0mg/l
DO final = 2.0mg/l
we know
BOD = [DO inital -DO final] * distribution factor
230 = [8 - 2] D.F
Distribution factor P 
Distribution factor P = =38.33
THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is
distribution factor 

V = 7.826 ml
is the volume of the sample when the water content is 10%.
<u>Explanation:</u>
Given Data:

First has a natural water content of 25% =
= 0.25
Shrinkage limit, 

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
![V \propto[1+e]](https://tex.z-dn.net/?f=V%20%5Cpropto%5B1%2Be%5D)
------> eq 1

The above equation is at
,

Applying the given values, we get

Shrinkage limit is lowest water content

Applying the given values, we get

Applying the found values in eq 1, we get

