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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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Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0
3 years ago
Read 2 more answers
A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =
oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0
3 years ago
What is the answer ?
tigry1 [53]
Is the control group
4 0
3 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
Anettt [7]

Answer:

Q=4.98\times 10^{-3}\ m^3/s

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

Pressure difference given as

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s

7 0
3 years ago
How could you nonverbally express interest while speaking to a customer in person? Check all of the boxes that apply.
JulijaS [17]

Answer: B,D,E, and G or 2,4,5,7

Explanation: Hope this helps Trust the answer I took the test!

5 0
3 years ago
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