The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
The potential difference across 3 Ohm resistor is 20V.
The resistors are connected in parallel which means all the three resistances have a fully potential difference of 20V.
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:
![l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft](https://tex.z-dn.net/?f=l%3D%5Cfrac%7B3-2%7D%7B2%7D%5C%2Cft%2B2%5C%2Cft%5C%5Cl%3D2.5%5C%2Cft)
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:
![A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}](https://tex.z-dn.net/?f=A%3Dd%5Ctimes%20l%5C%5CA%3D5%5C%2Cft%20%5Ctimes%202.5%5C%2C%20ft%5C%5CA%3D12.5%5C%2C%20ft%5E%7B2%7D)
And the raising speed <em>v </em>of the water is given by:
![v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bq%7D%7BA%7D%5C%5Cv%3D%5Cfrac%7B1%5C%2C%20%5Cfrac%7Bft%5E3%7D%7Bmin%7D%7D%7B12.5%5C%2C%20ft%5E2%7D%5C%5Cv%3D0.08%5C%2C%20%5Cfrac%7Bft%7D%7Bmin%7D)
where <em>q</em> is the water flow (1 cubic foot per minute).
106.68 centimetres are in 3.50 feet
Answer:
3.675 m
Explanation:
![v_{yo}=0](https://tex.z-dn.net/?f=v_%7Byo%7D%3D0)
X-direction | Y-direction
| ![y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%2Bv_%7Byo%7Dt%2B%5Cfrac%7B1%7D%7B2%7Da_%7By%7Dt%5E2)
|![y=0+0+\frac{1}{2} (9.8)(0.75)](https://tex.z-dn.net/?f=y%3D0%2B0%2B%5Cfrac%7B1%7D%7B2%7D%20%289.8%29%280.75%29)
| ![y=3.675 m](https://tex.z-dn.net/?f=y%3D3.675%20m)
Hope it helps