A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co
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Answer:
velocity = 62.89 m/s in 58 degree measured from the x-axis
Explanation:
Relevant information:
Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.
Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.
Before collision Momentum of A = 1000 x 100 = kg - m/s in the right direction.
Before collision Momentum of B = 2000 x 80 = 1.6 x kg - m/s in upward direction.
Mass of System of after collision = 1000 + 2000 = 3000 kg
Now applying the Momentum Conservation, we get
Initial momentum in right direction = final momentum in right direction =
And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x
So, =
m/s
and m/s
Therefore, velocity is =
=
= 62.89 m/s
And direction is
tan θ = = 1.6
therefore,
= from x-axis
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