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kifflom [539]
3 years ago
10

A 25 lb sacrificial Mg anode is attached to the steel hull of a container ship. If the anode completely corrodes within 3 months

, what is the average current produced by the anode?
Engineering
1 answer:
irakobra [83]3 years ago
3 0

Answer:

The average current will be of 6.36 A.

Explanation:

The anode capacity of magnesium is C = 550 A*h/lb

A month has 30 days.

A day has 24 hours.

Therefore 3 months have:

t = 3 * 30 * 24 = 2160 hours.

The average current is then:

I = C * m / t

I = 550 * 25 / 2160 = 6.36 A

The average current will be of 6.36 A.

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(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
olya-2409 [2.1K]

Answer:

The answer is "4.35 \times 10^{-3}\  mm and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point  

y =  yo + \frac{k}{\sqrt{x}}

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.  

Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

People can get yo = 275 MPa with both equations and k= 15.5 Mpa mm^{\frac{1}{2}}.

To substitute the answer,  

310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\  mm

In point b:

The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

equation is:

75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

5 0
3 years ago
Which website suffixes are usually the least credible? Check all that apply.
zmey [24]
.com hope this helps
7 0
3 years ago
Read 2 more answers
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

4 0
3 years ago
A keyboard, monitor and mouse are all considered as computer hardware and a program is known as a software.
Iteru [2.4K]

Answer:

This should be true because program is a software and mouse and keyboard are specific computer hardware to make your exsperience with a computer more efficent and faster.

5 0
3 years ago
Explain the following statement: In a windowless proportional counter, the output pulse height from an alpha particle source wil
-BARSIC- [3]

Answer / Explanation:

Regarding α and β Particles in Windowless Counter, the range of α particles is lower than β particles. Alpha particles typically have range less than the dimensions of the gas chamber so that proportional  counters are able to easily record. Hence,  with almost 100% efficiency, each particle which enters the so called active volume.

Then, since the pulse height spectra depends on the number of ion pairs which have formed,  an aplha particle with higher energy creates more ion pairs in the chamber.  However, beta particle range usually exceeds the dimensions of the chamber and therefore  most of the betas hit the walls where they deposit energy. Then, fewer ion pairs are formed  because very few β’s give their energy to the bulk gas.

5 0
3 years ago
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