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2 years ago

What is the first thing a person must do before anything can be done today corroded area

Engineering

1 answer:

melamori03 [73]2 years ago

8 0

Answer:

Explanation:

We have learned that three things are required for the anodic and cathodic steps of corrosion to occur: an electrolyte, an exposed metal surface, and an electron acceptor. It follows, then, that we can prevent corrosion by removing one of these essential conditions.

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Answer every question of this quiz

Reil [10]

I'd say number 4, number 3 looks like an exhaust valve

5 0

4 years ago

Suzanne Brett wants to borrow $55,000 from the bank. The interest rate is 6.5% and the term is for 5 years.

Otrada [13]

Answer: $14575

$55000

6.5%

5 years

Total Payment Amount: $72875

Yearly payment :$72875/5= $14575

8 0

2 years ago

What kind of robot should i make

schepotkina [342]

Answer:

It really depends on what you want it to do. I would make one that does chores around the house so I don't have to.

Explanation:

4 0

2 years ago

The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi

coldgirl [10]

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, $W=p(v_{2}-v_{1})$

On putting the estimated values, we get

⇒ $=200000(2-6)$

⇒ $=200000\times (-4)$

⇒ $=800,000 \ N.m$

Now,

Gas ideal equation will be:

⇒ $pv_{1}=mRT_{1}$

On putting the values. we get

⇒ $200000\times 6=1.5\times 2077\times T_{1}$

⇒ $T_{1}=\frac{1200000}{3115.5}$

⇒ $=385.1^{\circ}K$ (Initial temperature of helium)

and,

⇒ $pv_{2}=mRT_{2}$

On putting the values, we get

⇒ $200000\times 2=1.5\times 2077\times T_{2}$

⇒ $T_{2}=\frac{400000}{3115.5}$

⇒ $=128.3^{\circ}K$ (Final temperature of helium)

3 0

3 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago