What is the half-life of radium-226?
2 answers:
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The current flowing in silicon bar is 2.02 10^-12 A.
<u>Explanation:</u>
Length of silicon bar, l = 10 μm = 0.001 cm
Free electron density, Ne = 104 cm^3
Hole density, Nh = 1016 cm^3
μn = 1200 cm^2 / V s
μр = 500 cm^2 / V s
The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.
J = Je + Jh
J = n qE μn + p qE μp
where, n and p are electron and hole densities.
J = Eq (n μn + p μp)
we know that E = V / l
So, J = (V / l) q (n μn + p μp)
J = (1.6 10^-19) / 0.001 (104
1200 + 1016
500)
J = 1012480 10^-16 A / m^2.
or
J = 1.01 10^-9 A / m^2
Current, I = JA
A is the area of bar, A = 20 μm = 0.002 cm
I = 1.01 10^-9
0.002 = 2.02
10^-12
So, the current flowing in silicon bar is 2.02 10^-12 A.
Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = m
Width of the window = m
Area of the window =
Difference in air pressure = Inside pressure - Outside pressure
= atm =
atm
Conversion of the pressure in its SI unit.
⇒ atm =
Pa
⇒ atm =
Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒
⇒
⇒ Plugging the values.
⇒
⇒ Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
Answer:
25 N
Explanation:
Work is a product of force and perpendicular distance moved.
W=Fd where F is force exerted and d is perpendicular distance.
However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as
Therefore,
Making F the subject of the formula then
where
is the angle of inclination. Substituting 190 J for W then 18 degrees for
and 8 m for d then