Answer:
add resistance to a circuit
Explanation:
It depends on the design in which it is incorporated. A fixed resistor has many uses, including, but not limited to ...
- dropping voltage
- limiting current
- contributing to a time delay
- adjusting frequency response
- eliminating (or creating) signal reflections
- acting as a fuse
- calibrating or trimming a response
- providing protection against electrical shock or ESD
- acting as a reference when measuring variable resistors
Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= 
By putting values, we get
= 
= 
= 
hence,
The average force will be:
= 
= 
= 
Or,
= 
Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
Answer:
Amount of air left in the cylinder=m
=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2
Initial temperature=T
=20 C
Final Volume=
=0.1 
Using gas equation

m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.

Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0