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dybincka [34]
3 years ago
11

Which of the following explains how electricity flows?

Physics
1 answer:
Yuri [45]3 years ago
7 0
Electrons flow from the positive end of a source towards the negative end
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Jesse celebrated his birthday yesterday by eating a giant birthday cake with 17 candles. What type of energy conversion ocurred
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Answer:

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Thermal to chemical

6 0
3 years ago
Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem
Tresset [83]

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

7 0
3 years ago
An iron wire has a cross-sectional area equal to 5.00×10⁻⁶ m² . Carry out the following steps to determine the drift speed of th
Doss [256]
  1. In mass, there are 55.85 × 10⁻³ kg/mol in in 1 mole of iron.
  2. The molar density of iron is equal to 1.41 × 10⁵ mol/m³.
  3. The density of iron atoms is equal to 8.49 × 10²⁸ atoms/m³.
  4. The number density of conduction electrons is equal to 1.70 × 10²⁹ conduction electrons/m³.
  5. The drift speed of conduction electrons is equal to 2.21 × 10⁻⁴ m/s.

<h3>How to calculate the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = (m × σ × V)/ρ × e × f × l)

V = I/(n × A × Q)

Where:

  • U represents the drift speed of the conduction electrons, in m/s.
  • m represents the molecular mass of the metal, in kg.
  • e represents the elementary charge, in C.
  • f represents the number of free electrons per atom.
  • σ represents the electric conductivity of the medium at a particular temperature in S/m.
  • ρ represents the density of the conductor, in kg/m³.
  • ℓ represents the length of the conductor, in m.
  • ΔV represents the voltage applied or potential difference across the conductor in V.

<h3>How many kilograms are there in 1 mole of iron? </h3>

Molar mass of iron = 55.85 g/mol.

In Kilograms, we have:

Mass = 55.85 × 1/1000

Mass = 55.85 × 10⁻³ kg/mol.

For the molar density of iron, we have:

Molar density = density/molar mass

Molar density = 7874/0.056

Molar density = 1.41 × 10⁵ mol/m³.

For the density of iron atoms, we have:

Density of iron atoms = Avogadro's constant × molar density

Density of iron atoms = 6.023 × 10²³ × 1.406 × 10⁵

Density of iron atoms = 8.49 × 10²⁸ atoms/m³.

For the number density of conduction electrons, we have:

Fe ---> Fe²⁺ + 2e⁻

Number density of conduction electrons = 2 conduction electrons/1 atom of iron

Number density of conduction electrons = 2 × 8.49 × 10²⁸

Number density of conduction electrons = 1.70 × 10²⁹ conduction electrons/m³.

For the drift speed of conduction electrons, we have:

V = I/(n × A × Q)

V = 30/(1.70 × 10²⁹ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁶)

Drift speed, V = 2.21 × 10⁻⁴ m/s.

Read more on drift speed here: brainly.com/question/15219891

#SPJ4

Complete Question:

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).

(c) Calculate the number density of iron atoms using Avogadro’s number.

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.

(e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

4 0
2 years ago
A 243 mL cup of whole milk contains about 45 mg of cholesterol. Express the cholesterol concentration of the milk in kilograms p
timama [110]

Answer: 0.1851 \frac{kg}{m^{3}}

Explanation:

We have this concentration in units of mg/ml:

\frac{45 mg}{243 ml}

And we need to express it in kg/m^{3}, knowing:

1 ml= 1 cm^{3}

1 m^{3}= (100 cm) ^{3}

1g = 1000 mg

1 kg = 1000 g

Hence:

\frac{45 mg}{243 cm^{3}} \frac{1 g}{1000 mg} \frac{1 kg}{1000 g} \frac{(100 cm) ^{3}}{1 m^{3}}=0.1851 \frac{kg}{m^{3}}

5 0
4 years ago
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