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3 years ago

PLEASE HELP WILL GIVE 25 POINTS!!!!!

Physics

2 answers:

zimovet [89]3 years ago

8 0

The answer is B. 1,4, and 6

icang [17]3 years ago

5 0

Answer : The correct option is, (B) 1, 4, and 6

Explanation :

Boyle's Law : It is defined as the pressure is inversely proportional to the volume of the gas at constant temperature and number of moles. In this volume of gas changes.

$P\propto \frac{1}{V}$

Charles' Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles. In this volume of gas changes.

$V\propto T$

Gay-Lussac's Law : It is defined as the pressure is directly proportional to the temperature of the gas at constant volume and number of moles. In this volume of gas is constant.

$P\propto T$

Therefore, the volume should be placed in cell numbered 1, 4, and 6.

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Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

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Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

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