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4 years ago

Which substance is used to remove rust from metal?

Physics

2 answers:

Vlad [161]4 years ago

7 0

Answer : The correct option is, (B) Hydrochloric acid(HCl)

Explanation :

Rust is an iron oxide, which is formed by the reaction of iron with oxygen in the presence of water.

As we know that the rust is an iron oxide (base) which is removed by an acid.

As per given options, the acid used to remove the rust from the metal is hydrochloric acid and not sulfuric acid because sulfuric acid is an oxidizing agent which oxidizes the metal more.

While the other options, ammonia and sodium hydroxide are a base.

Hence, hydrochloric acid is used to remove rust from metal.

Andreyy894 years ago

3 0

Hydrochloric acid would be the correct answer

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A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t

Mandarinka [93]

Answer:

The mmf required is $1.125$ × $10^{-3}$ A

Explanation:

The Magnetomotive force (mmf) is given by the formula below

$F_{M} = Hl\\$

where $F_{M}$ is the Magnetomotive force (mmf)

$H$ is the Magnetic field strength

$l$ is the magnetic length

The magnetic permeability μ is given by

μ = $B / H$

Where $B$ is the Magnetic flux density

and $H$ is the Magnetic field strength

From the question,

$B$ = 1.2Wb/m^2

μ = 1600m

From μ = $B / H$

∴ $H = B/$ μ

$H = 1.2 / 1600\\$

$H = 7.5$ × $10^{-4}$ A/m

Now, for the Magnetomotive force (mmf)

$F_{M} = Hl\\$

From the question

$l$ = 1.5 m

∴ $F_{M} = 7.5$ × $10^{-4}$ × $1.5$

$F_{M} = 1.125$ × $10^{-3} A$

Hence, The mmf required is $1.125$ × $10^{-3}$ A

3 0

3 years ago

A hiking trail is 150 meters long. A mountain lion can run the entire trail in 30 seconds.

konstantin123 [22]

Answer:

5 meters per second

explanation

sana po makatulong

7 0

3 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago

A rectangular reflecting pool is 56.5 ft wide and 141 ft long. What is the area of the pool in square meters

konstantin123 [22]

By definition, the area of the pool is 740.112 m².

<h3>Definition of area</h3>

The area is the measure of a space delimited by a contour which is called perimeter.

That is, the area gives an idea of how much surface a figure covers and is defined as the measure of the region or surface enclosed by a geometric figure.

<h3>Area of a rectangle</h3>

A rectangle is a flat geometric figure with four sides, of which two sides that are opposite parallel to each other have the same length and the remaining two have another length.

The area of a rectangle is calculated by multiplying the lengths of two contiguous sides of the figure which are different from each other, that is, the length is multiplied by the width:

Area= length× width

A rectangular reflecting pool is 56.5 ft wide and 141 ft long. So, the area of the rectangular pool is calculated as:

Area= length× width

Area= 141 ft× 56.5 ft

Solving:

<u><em>Area= 7966.5 ft²= 740.112 m²</em></u> (being 1 ft²= 0.092903 m²)

So, the area of the pool is 740.112 m².

Learn more about the area of a rectangle:

brainly.com/question/11202023

brainly.com/question/13048427

4 0

3 years ago

What are the three most influential variables that affect electrical force?

snow_tiger [21]

I know that the answer is 3 because 1+2 is 3

4 0

4 years ago