To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as
Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is
At the same time since Newton's second law we know that:
Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to
Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:
But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:
Therefore there is a weight loss of 0.3N every kilometer.
Answer:
answer is option 3
Explanation:
because you need to find out time
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.
