An upright spring with a 96g mass on it is compressed 2 cm. When
1 answer:
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Answer:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
- Speed of the object
- Cross-sectional area of the object
- Shape of the object
Formula:
As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.
Answer:
0.08 ft/min
Explanation:
To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.
So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:
since the difference between the upper and lower base is the increase in the base and we are only at halft the height.
Now we can calculate the longitudinal section <em>A</em> at that point:
And the raising speed <em>v </em>of the water is given by:
where <em>q</em> is the water flow (1 cubic foot per minute).