The volume of a cube is given by:

where L is the measure of each side (which corresponds to its width, since all sides of a cube are equal).
Plugging the data of the problem, L=32.1 cm, inside the equation, we find the volume:

A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
Answer:
2.5 m/s
Explanation:
The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground
v = 0 m/s + 2.5 m/s
v = 2.5 m/s
At the moment it is released, the package is rising at 2.5 m/s.
The correct answer is:
<span>Point charges must be in a vacuum.
In fact, the usual form for of the Coulomb's law is:
</span>

<span>where
</span>

is the permittivity of free space
<span>q1 and q2 are the two charges
q is the separation between the two charges
However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
</span>

where

is the relative permittivity, which takes into account the dielectric effects of the material.