Answer:
part a : <em>The dry unit weight is 0.0616 </em>
<em />
part b : <em>The void ratio is 0.77</em>
part c : <em>Degree of Saturation is 0.43</em>
part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>
Explanation:
Part a
Dry Unit Weight
The dry unit weight is given as

Here
is the dry unit weight which is to be calculated- γ is the bulk unit weight given as

- w is the moisture content in percentage, given as 12%
Substituting values

<em>The dry unit weight is 0.0616 </em>
<em />
Part b
Void Ratio
The void ratio is given as

Here
- e is the void ratio which is to be calculated
is the dry unit weight which is calculated in part a
is the water unit weight which is 62.4
or 0.04 
- G is the specific gravity which is given as 2.72
Substituting values

<em>The void ratio is 0.77</em>
Part c
Degree of Saturation
Degree of Saturation is given as

Here
- e is the void ratio which is calculated in part b
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Substituting values

<em>Degree of Saturation is 0.43</em>
Part d
Additional Water needed
For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

Here
is the zero air unit weight which is to be calculated
is the water unit weight which is 62.4
or 0.04 
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction

Now as the volume is known, the the overall weight is given as

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.
Answer:
100/10 = 10 , 10 × 10 = 100÷20 = 5
I'm pretty sure its wrong
Answer:
By 16.7% or 0.167 IPM
Explanation:
Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :

m is the mass of object


m = 0.25 kg


k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer: 10 and 35 degrees
Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment