Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that it starts from rest and moves on runway by total distance 165 m
so we will have



now we have

Part b)
Now for take off time we will have



Part c)



Answer:
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Explanation:
the fundamental answer is without regular supervision of definition of weights and measures,commerce exchange will be impossible and there would be no market whatsoever for anything.
<h3><u>Answer;</u></h3>
= 480 Joules
<h3><u>Explanation;</u></h3>
We use the formula, Q - W = ΔU
Where, Q = Heat transferred to the system
W = Work done by the system
ΔU = Change of internal energy.
As per the question, Q = 658 J
ΔU = 178 J
Thus, W = Q - ΔU = (658 - 178) J = 480 J.
The energy used to do work by the system is 480 J.
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.