Noble gases are not highly reactive
By using drift velocity of the electron, the current flow is 7.20 ampere.
We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as
v = I / (n . A . q)
where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.
From the question above, we know that
d = 2.097 mm
r = (0.002097 / 2) m
v = 1.54 mm/s = 0.00154 m/s
ρ = 8.92 x 10³ kg/m³
q = e = 1.6 x 10¯¹⁹C
Find the atom density
n = Na x ρ / Mr
where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).
n = 6.022 x 10²³ x 8.92 x 10³ / 0.635
n = 8.46 x 10²⁷ /m³
Find the current flows
v = I / (n . A . q)
0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)
0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)
I = 7.20 ampere
For more on drift velocity at: brainly.com/question/25700682
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Answer:
Explanation:
The inlet specific volume of air is given by:
The mass flow rates is expressed as:
The energy balance for the system can the be expresses in the rate form as:
![E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}](https://tex.z-dn.net/?f=E_%7Bin%7D-E_%7Bout%7D%3D%5Cbigtriangleup%20%5Cdot%20E%3D0%5C%5C%5C%5CE_%7Bin%7D%3DE_%7Bout%7D%5C%5C%5C%5C%5Cdot%20m%28h_1%2B0.5V_1%5E2%29%3D%5Cdot%20W_%7Bout%7D%2B%5Cdot%20m%28h_2%2B0.5V_2%5E2%29%2BQ_%7Bout%7D%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D%5Cdot%20m%28h_2-h_1%2B0.5%28V_2%5E2-V_1%5E2%29%29%3D-m%28%7Bcp%28T_2-t_1%29%2B0.5%28V_2%5E2-V_1%5E2%29%7D%29%5C%5C%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D-%2810.42lbm%2Fs%29%5B%280.25%5Cfrac%7BBtu%7D%7Blbm.%5Ctextdegree%20F%7D%29%28300-900%29%5Ctextdegree%20F%2B0.5%28%28700ft%2Fs%29%5E2-%28350ft%2Fs%29%5E2%29%28%5Cfrac%7B1%5Cfrac%7BBtu%7D%7Blbm%7D%7D%7B25037ft%5E2%2Fs%5E2%7D%29%5D%5C%5C%5C%5C%5C%5C%5C%5C%3D1486.5%5Cfrac%7BBtu%7D%7Bs%7D)
Hence, the mass flow rate of the air is 1486.5Btu/s
A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.
England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.
For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.
The catch in observing the eclipse is:
<em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.
Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.
This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".
