The definition of Force
2 answers:
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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 =
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
<h3><u>Mass of an object affect the outcome of unbalanced forces:</u></h3>
Newton’s second law of motion deals with the result of motion of an object when unbalanced forces are applied. The second law of motion establish the relationship between mass, acceleration and unbalanced forces of the object. The below points explains the relation between unbalanced forces acting on mass of the object. The acceleration of object would be higher when the unbalanced force is higher.
The equation for the unbalanced force is,
For Example: Two masses of 500 kg and 1 kg is applied with same unbalanced forces. The change in motion of 500 kg would be very much less than the change in motion of 1 kg.
Answer:
a) please find the attachment
(b) 3.65 m/s^2
c) 2.5 kg
d) 0.617 W
T<weight of the hanging block
Explanation:
a) please find the attachment
(b) Let +x be to the right and +y be upward.
The magnitude of acceleration is the same for the two blocks.
In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:
∑Fx=ma_x
T=m1a_x
14.7=4.10a_x
a_x= 3.65 m/s^2
c) <em>in order to calculate m we will apply newton second law on the hanging </em>
<em> block</em>
<em> </em>∑F=ma_y
T-W= -ma_y
T-mg= -ma_y
T=mg-ma_y
T=m(g-a_y)
a_x=a_y
14.7=m(9.8-3.65)
m = 2.5 kg
<em>the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax</em>
d) calculate the weight of the hanging block :
W=mg
W=2.5*9.8
=25 N
T=14.7/25
=0.617 W
T<weight of the hanging block
Explanation:
hope this helps you
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