Two thousand pieces will flow through from the first machine A to the final machine F based on the given sequence of operations.
1 answer:
The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
<em>"Your</em><em> </em><em>question is not complete, it seems to be missing the following information;"</em>
The sequence of operation is A - E - D - C - B - A - F
The given parameters;
- <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
- <em>initial unit load specified in the first machine, L₁ = 50</em>
- <em>final unit load, L₂ = 100 </em>
- <em>the capacity of the vehicle = 1 unit load</em>
<em />
The given sequence of operation of the vehicle;
A - E - D - C - B - A - F
<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>
The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.
100 unit loads ----------------- 6 trips
2000 unit loads --------------- ?
Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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Answer:
The correct answers are:
a. % w = 33.3%
b. mass of water = 45g
Explanation:
First, let us define the parameters in the question:
void ratio e = =
Specific gravity =
% Saturation S = ×
=
×
water content w = =
a) To calculate the lower and upper limits of water content:
when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.
when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.
Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.
To get the relationship between water content and saturation, we will manipulate the equations above;
w =
Recall; mass = Density × volume
w =
From eqn. (2) =
∴
putting eqn. (6) into (5)
w =
Again, from eqn (1)
substituting into eqn. (7)
∴
With eqn. (7), we can calculate
upper limit of water content
when S = 100% = 1
Given,
∴
∴ %w = 33.3%
Lower limit of water content
when S = 1% = 0.01
∴ % w = 0.33%
b) Calculating mass of water in 100 cm³ sample of soil ( )
Given, , S = 50% = 0.5
%S = ×
=
×
0.50 =
mass of water =