A student jobs around a square park two times. Starting and ending at the gate to the park. The square jogging track is 40 meter
2 answers:
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Answer:
Approximately (rounded down,) assuming that
.
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules, ):
.
Height of the weight (should be in meters, ):
.
Energy required to lift the weight by without acceleration:
.
At an efficiency of , the actual amount of energy required to raise this weight to that height would be:
.
Divide by
to find the number of times this weight could be lifted up within that energy budget:
.
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.
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 =
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 =
v =
v = 3.22 x 10⁻⁴ m/s
Therefore, the drift electron is 3.22 x 10⁻⁴ m/s