It is weigh-in time for the local under 85 kg rugby team. The bathroom scale that is used to assess eligibility can be described
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To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:
Where,
G = Gravitational Universal Constant
M = Mass of Planet
r = Radius of the planet ('h' would be the orbit from the surface)
The escape velocity is
Through this equation we can find the mass of the Planet in function of the distance, therefore
The orbital velocity is
The time period of revolution is,
Therefore the orbital period of the satellite is closes to 1 hour and 12 min
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.