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:
"where crests and troughs have their maxima at the same time"
Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=<span>1.67 10^-27 kg
</span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5= - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV
the final answer is </span><span>Eb /nucleon </span><span>= -5.625 x10^35 eV</span>
Answer:
Electric field acting on the electron is 127500 N/C.
Explanation:
It is given that,
Mass of an electron,
Charge on electron,
Initial speed of electron, u = 0
Final speed of electron,
Distance covered, s = 2 cm = 0.02 m
We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :
According to Newton's law, force acting on the electron is given by :
F = ma
Electric force is given by :
F = q E, E = electric field
E = 127500 N/C
So, the electric field is 127500 N/C. Hence, this is the required solution.
Answer
given,
angle between two polarizing filters = 45°
filter reduce intensity = ?
a) I = I₀ Cos² θ
here θ = 45⁰
intensity of the light is reduced by 0.500
correct answer from the given option D
b) direction of the polarization
θ = 45°