Answer:
a)n= 3.125 x 
electrons.
b)J= 1.515 x 
A/m²
c)
=1.114 x 
m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x 
m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x 
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x 
C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
= 
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
Answer:Ultraviolet radiation has shorter wavelengths and higher energy than infrared radiation.
Explanation: Electromagnetic radiation radiations which have both electrical and magnetic properties,they can be transmitted through space or through a medium.
It includes Gamma radiation, infra-red, visible light, Ultraviolet radiation etc they occur with different wavelength, the lower the wavelength the higher the Energy dissipated per photon. According to their order of decreasing wavelength and increased energy they are classified as follows.
RADIO WAVE, MICRO WAVE, INFRA-RED, VISIBLE LIGHT, ULTRAVIOLET RAY, X-RAY, GAMMA RAYS.
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
<u>Given the following data;</u>
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;
Where;
- P is the resistivity of the material.
- L is the length of the material.
- A is the cross-sectional area of the material.
First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;
<em>Resistance = 9.95 Ohms </em>
Answer:
D. The period would decrease by sqrt (2)
Explanation:
The period of a mass-spring system is given by:
where
m is the mass
k is the spring constant of the spring
If the spring constant is doubled,
k' = 2k
So the new period will be
So the correct answer is
D. The period would decrease by sqrt (2)
