Answer:
They can be seen from a distance of 4.372 kilometers.
Explanation:
Using the Reyligh creterion for diffraction through a circular aperture we have
where symbol's have their usual meaning
thus applying values we get


Answer:
I Think its ABC
Explanation:
No air is the lightest
Helium is lighter than regular air that's why it goes up so then the regular air would be heaviest.
Answer: 1.28 sec
Explanation:
Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:
V = ∆x / ∆t
Solving for ∆t, we have:
∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution =
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:

Plug in the given values and solve for
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E = 
indicates that the field is E for r = 2m
E =
(1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C