Complete Question
A 10 gauge copper wire carries a current of 20 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.)
mm/s
Answer:
The drift velocity is 
Explanation:
From the question we are told that
The current on the copper is 
The cross-sectional area is 
The number of copper atom in the wire is mathematically evaluated
Where 
is the density of copper with a value 
is the Avogadro's number with a value 
Z is the molar mass of copper with a value 
So
Given the 1 atom is equivalent to 1 free electron then the number of free electron is
The current through the wire is mathematically represented as
substituting values
=>
Answer:
No. of Neutrons = 3
Explanation:
The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.
Mass Number = No. of Protons + No. of Neutrons = 6
Atomic Number = Number of Electrons = No. of Protons = 3
Therefore,
Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons
Mass Number - Atomic Number = No. of Neutrons
No. of Neutrons = 6 - 3
<u>No. of Neutrons = 3 </u>
To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2<span>°</span>)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.
Answer:
Rubber or plastic covers are bad conductors of electricity. So they do not allow the electric current to pass through it.
Explanation:
Rubber and plastic are bad conductors of electricity, therefore when handling a tool with a rubber handle, the electricity will not pass through it.
