Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in centim
1 answer:
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Answer:
<em>a) 17.05 mph</em>
<em>b) 54.7° northeast direction</em>
<em>c) 10.71 mph</em>
<em>The direction is -22.58° relative to the east.</em>
<em></em>
<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>
Explanation:
The question is a little confusing but, I guess the correct question should be;
You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.
a) What is your airspeed?
b) What angle (direction) are you flying?
c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?
NB: The difference in the question and my suggestion is highlighted boldly.
Your speed = 14 mph
direction is 45° northeast
Th wind speed = 4 mph
direction is north
We resolve the your speed and the wind speed into the horizontal and vertical components
For vertical the component component
= 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph
For the horizontal speed component
= 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph
Resultant speed =
==> = <em>17.05 mph This is your airspeed</em>
b) To get your direction, we use
tan ∅ = /
tan ∅ = 13.89/9.89 = 1.413
∅ = (1.413) = <em>54.7° northeast direction</em>
c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .
In this case,
= 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph
Resultant speed =
==> = <em>10.71 mph This is your airspeed</em>
Your direction will be,
tan ∅ = /
tan ∅ = -4.11/9.89 = -0.416
∅ = (-0.416) =<em> -22.58° this is the angle you'll travel relative to the east.</em>
<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = / ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.