Anyone really , but you must be enrolled in an agriculture course at your school.
Answer:
Questions are answered in the explanation section.
Explanation:
a) Of both metals, lead will corrode, since it reacts with seawater to form oxides, chlorides, etc., while bronze is more resistant to oxidation.
b) The corrosion rate will remain constant, because the concentration of seawater does not change over time, therefore, the corrosion rate is independent of the area that is exposed to seawater.
c) Corrosion can be reduced by using protective coatings, so that the metal is isolated from sea water, for example, paints, acrylic powders, etc.
Answer: N has to be lesser than or equal to 1666.
Explanation:
Cost of parts N in FPGA = $15N
Cost of parts N in gate array = $3N + $20000
Cost of parts N in standard cell = $1N + $100000
So,
15N < 3N + 20000 lets say this is equation 1
(cost of FPGA lesser than that of gate array)
Also. 15N < 1N + 100000 lets say this is equation 2
(cost of FPGA lesser than that of standardcell)
Now
From equation 1
12N < 20000
N < 1666.67
From equation 2
14N < 100000
N < 7142.85
AT the same time, Both conditions must hold true
So N <= 1666 (Since N has to be an integer)
N has to be lesser than or equal to 1666.
Answer:
F = 2840.3 N
Explanation:
Given:
- Diameter of window D = 0.3 m
- Midpoint of window from sea level h = 4 m
- Specific gravity of sea water S.G = 1.024
- Density of water p = 1000 kg/m^3
Find:
The hydro-static force F_r acting on the mid-point of the window.
Solution:
- The average pressure P acting on the midpoint of the window:
P = S.G p*g*h
P = 1.024*1000*9.81*4
P = 40181.76
- The hydro-static force F_r acting on the mid-point of the window:
F = P*A = P*pi*D^2 / 4
F = 40181.76*pi*0.3^2 / 4
F = 2840.3 N
Parallel hookups increase amp hour capacity but voltage remains the same... 1.5 volts
Series the two and voltage is 3.0 volts.