4. E
5. D
6. F
Hope this helps
Between magnitude of the average 4sec
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>
Vo = 18 m/s
angle 35 degrees
1) Components of the initial velocity
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s
2) Equations of postion:
x = Vox*t
y = Voy*t - gt^2 / 2
3) Calculations
A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s
x = 14.74 * t
t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m
t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m
t = 1.5s => x = 22.11 m
t = 2s => x = 29.48 m
B)
y = Voy*t - gt^2 / 2
Voy = 10.32 m/s
g = 10 m/s (approximation)
y = 10.32*t - 5t^2
t = 0.5 s=> y = 3.91m
t = 1 s => y = 5.32m
t = 1.5 s => y = 4.23m
t = 2 s => y = 0.64 m
Answer:
The room become bright in the day by a process call scattering of light or Rayleigh scattering.
Explanation:
It is called Rayleigh scattering or scattering of light because Rayleigh scattering is the scattering if light or electromagnetic radiation by smaller particles which have radius of less than 110 nanometer in a medium and the wavelengths of the electromagnetic. Wave or light remain unchanged. The scattering of light occur in the day time in the room and this bring of brighten up of the room.
