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
Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Its 0.001
0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m
I think because if you’ve already turned it in they might as well grade asap instead of waiting
The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.
<u>Explanation:</u>
- Since
> 
because 
> Vt. - By the saturation region the MOSFET is operating.
- A specific source voltage and gate of NMOS, the voltage get drained during the specific level, the drain voltage is rises beyond where there is no effect of current during saturated region.
- MOSFET is a transistor which is a device of semiconductor vastly used for the electronic amplifying signals and switching in the devices of electronics.
- The core of this is integrated circuit.
- It is fabricated and designed in an individual chips due to tiny sizes.
