Answer:
10
Explanation:
i = 5/.5 = 10 Amps. Hope this helps :)
3 protons should be your answer
Answer:
v = 12.4 [m/s]
Explanation:
With the speed and Area information, we can determine the volumetric flow.
where:
r = radius = 0.0120 [m]
v = 2.88 [m/s]
![A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2A%280.0120%29%5E%7B2%7D%20%5C%5CA%3D4.523%2A10%5E%7B-4%7D%20%5Bm%5D%5C%5C)
Therefore the flow is:
![V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]](https://tex.z-dn.net/?f=V%3D2.88%2A4.523%2A10%5E%7B-4%7D%20%5C%5CV%3D1.302%2A10%5E%7B-3%7D%20%5Bm%5E%7B3%7D%2Fs%20%5D)
Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.
![v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]](https://tex.z-dn.net/?f=v%3DV%2FA%5C%5Cv%3D1.302%2A10%5E%7B-3%7D%20%2F1.05%2A10%5E%7B-4%7D%20%5C%5Cv%3D12.4%5Bm%2Fs%5D)
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
