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3 years ago

How magnets are used in loud speakers

1 answer:

5 0

The electromagnet and the permanent magnet -- interact with each other as any two magnets do. The positive end of the electromagnet is attracted to the negative pole of the permanent magnetic field, and the negative pole of the electromagnet is repelled by the permanent magnet's negative pole

You might be interested in

Which 10-milliliter sample of water has the greatest degree of disorder?A)H2O(g) at 120°CB)H2O() at 80°CC)H2O() at 20°CD)H2O(s)

AnnyKZ [126]

Answer:

A) H₂O at 120°C

Explanation:

It is possible to think the higher temperature, the greatest degree of disorder. That is because with a high temperature, vibrations of molecules increases.

In general, at low temperatures, the molecules are in solid state (The lowest degree of disorder), increasing its temperature, molecules becomes in liquids, and, with more temperature, are gases (The greatest degree of disorder).

Thus, the sample that has the greatest degree of disorder is:

6 0

3 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average power of approximately 8.00

Archy [21]

Energy transferred per second in given area:
8000 x 3.9 / 10000
Energy transferred = 3.12 J

Using Planck's equation:
E = nhc/λ
n = (3.12 x 510 x 10⁻⁹) / (3 x 10⁸ x 6.63 x 10⁻³⁴)
n = 8 x 10¹⁸ photons

3 0

2 years ago

Gold forms face-centered cubic crystals. The atomic radius of a gold atom is 144 pm. Consider the face of a unit cell with the n

kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

X is the length of an edge of this unit cell

R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m

X = 144 X 10⁻¹²√8

X = 407.294 X 10⁻¹² m

X = 407.294 pm

Therefore, the length of an edge of this unit cell is 407.294 pm

8 0

2 years ago

6.) 50.0 mol H2O<br> ? molecules

8_murik_8 [283]

<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} )$ = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0

3 years ago

When CO2 decomposes into oxygen and carbon, it gives a gram ratio of 2.67:1 O2:C. When a 32.4g of CO2 decomposes, how many grams

Rufina [12.5K]

Answer : The mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

$CO_2\rightarrow O_2+C$

As we are given:

$\text{Mass of }O_2}:\text{Mass of }C=2.67:1{$

According to the law of conservation of mass,

Total mass of $CO_2$ = Mass of $O_2$ + Mass of C

Total mass of $CO_2$ = 2.67 + 1 = 3.67 g

Now we have to calculate the mass of $O_2$ and C.

$\text{Mass of }O_2=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }O_2=\frac{32.4g}{3.67g}\times 2.67=23.6g$

and,

$\text{Mass of }C=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }C=\frac{32.4g}{3.67g}\times 1=8.83g$

Therefore, the mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

6 0

3 years ago