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Rufina [12.5K]
3 years ago
14

The acceleration due to gravity vector is always in the ____ direction.

Physics
1 answer:
finlep [7]3 years ago
4 0
It goes in the downward direction
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in the hydrologic cycle, water from the ocean _____ (condenses, precipitates, transpires,or evaporates) into the atmosphere wher
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Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 k
tino4ka555 [31]

Answer:

\mathbf{ current(I) =1766.67 \ A}

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

13 \times  current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}

current(I) \times 10.92= 19292

current(I) = \dfrac{19292}{10.92}

\mathbf{ current(I) =1766.67 \ A}

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2 years ago
Which statements describe properties of stars? Check all that apply.
Shtirlitz [24]
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A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

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

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

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)

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

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

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

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

6 0
3 years ago
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