The answer would be 5 years
Answer:
Sheet metal forming dates back to the days of the blacksmith who used a hammer and hot oven to mold metal into the desired form. Today’s aircraft technician relies on a wide variety of powered and hand-operated tools to precisely bend and fold sheet metal to achieve the perfect shape. Forming tools include straight line machines, such as the bar folder and press brake, as well as rotary machines, such as the slip roll former. Forming sheet metal requires a variety of tools and equipment (both powered and manual), such as the piccolo former, shrinking and stretching tools, form blocks, and specialized hammers and mallets. [Figure 4-55]
Answer:
Explanation:
Pie charts generally should have no more than eight segments.
Answer:
A Bipolar Junction Transistor, or BJT, is a solid-state device in which the current flow between two terminals (the collector and the emitter) is controlled by the amount of current that flows through a third terminal (the base).
The main basic function of a BJT is to amplify current it will allow BJTs are used as amplifiers or switches to produce wide applicability in electronic equipment include mobile phones, industrial control, television, and radio transmitters. There are two different types of BJTs are available, they are NPN and PNP.
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 = 