Answer:
I believe the correct answer would be Weight.
Weight is a force that gravity exerts on an object.
Hope this helped :))))
The elevation difference between the PVC and the high point on the curve 0.1208 ft
<u>Explanation:</u>
g1 = 4.0%
= 0.04
The elevation of curve at a distance “x”from pvc is given by
![y=a x^{2}+b x+ ele p v c](https://tex.z-dn.net/?f=y%3Da%20x%5E%7B2%7D%2Bb%20x%2B%20ele%20p%20v%20c)
x = distance from pvc to point on curve
The highest point on vertical curve is at a distance "x" is given by
![= -g1L/g2-g1](https://tex.z-dn.net/?f=%3D%20%20-g1L%2Fg2-g1)
so, elevation of highest point ![y=a x^{2}+b x+ele p v c](https://tex.z-dn.net/?f=y%3Da%20x%5E%7B2%7D%2Bb%20x%2Bele%20p%20v%20c)
here,
![a=\frac{g_{2}-g_{1}}{2 L} \quad ; \quad b=g_{1} ; x=\frac{-g_{1} C}{g_{2}-g_{1}}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bg_%7B2%7D-g_%7B1%7D%7D%7B2%20L%7D%20%5Cquad%20%3B%20%5Cquad%20b%3Dg_%7B1%7D%20%3B%20x%3D%5Cfrac%7B-g_%7B1%7D%20C%7D%7Bg_%7B2%7D-g_%7B1%7D%7D)
substitute above values in "y" equation,
![y=\left(\frac{g_{2}-g_{1}}{2 C}\right) \cdot\left(\frac{-g_{1} L}{g_{2}-g_{1}}\right)^{2}+g_{1}\left(\frac{-g_{1} L}{g_{2}-g_{1}}\right)+\text { elepvc }](https://tex.z-dn.net/?f=y%3D%5Cleft%28%5Cfrac%7Bg_%7B2%7D-g_%7B1%7D%7D%7B2%20C%7D%5Cright%29%20%5Ccdot%5Cleft%28%5Cfrac%7B-g_%7B1%7D%20L%7D%7Bg_%7B2%7D-g_%7B1%7D%7D%5Cright%29%5E%7B2%7D%2Bg_%7B1%7D%5Cleft%28%5Cfrac%7B-g_%7B1%7D%20L%7D%7Bg_%7B2%7D-g_%7B1%7D%7D%5Cright%29%2B%5Ctext%20%7B%20elepvc%20%7D)
![\frac{\left(g_{2}-g_{1}\right)}{2 L} \cdot \frac{g_{1}^{\prime} L^{\prime}}{\left(g_{2}-g_{1}\right)}-\frac{g_{1}^{\prime} L}{\left(g_{2}-g_{1}\right)}+\text { ele pvc }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D%7B2%20L%7D%20%5Ccdot%20%5Cfrac%7Bg_%7B1%7D%5E%7B%5Cprime%7D%20L%5E%7B%5Cprime%7D%7D%7B%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D-%5Cfrac%7Bg_%7B1%7D%5E%7B%5Cprime%7D%20L%7D%7B%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D%2B%5Ctext%20%7B%20ele%20pvc%20%7D)
![y=\frac{g_{1}^{\prime} L}{2\left(g_{2}-g_{1}\right)}-\frac{g_{1}^{\prime} L_{1}}{\left(g_{2}-g_{1}\right)}+\text { ele pvc }](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bg_%7B1%7D%5E%7B%5Cprime%7D%20L%7D%7B2%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D-%5Cfrac%7Bg_%7B1%7D%5E%7B%5Cprime%7D%20L_%7B1%7D%7D%7B%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D%2B%5Ctext%20%7B%20ele%20pvc%20%7D)
![y=-\frac{1}{2} \frac{g_{1}^{r} L}{\left(g_{2}-g_{1}\right)}+\text { elepvc }](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bg_%7B1%7D%5E%7Br%7D%20L%7D%7B%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D%2B%5Ctext%20%7B%20elepvc%20%7D)
![\text { elepvc }-y=\frac{1}{2} \frac{g_{1}^{r} L}{\left(g_{2}-g_{1}\right)}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20elepvc%20%7D-y%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bg_%7B1%7D%5E%7Br%7D%20L%7D%7B%5Cleft%28g_%7B2%7D-g_%7B1%7D%5Cright%29%7D)
Algebric sum of grades = A = ![g_{2}-g_{1}](https://tex.z-dn.net/?f=g_%7B2%7D-g_%7B1%7D)
As per AASHTO,for V= 60 mph, K =151 based on Stopping sight distance
Minimum length of vertical curve
![L=K A\\](https://tex.z-dn.net/?f=L%3DK%20A%5C%5C)
![L=151 A](https://tex.z-dn.net/?f=L%3D151%20A)
![A=\frac{L}{151}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BL%7D%7B151%7D)
We know that,
![\text { ele pvc }-y=\frac{1}{2} \frac{g_{1}^{\prime} L}{A}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20ele%20pvc%20%7D-y%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bg_%7B1%7D%5E%7B%5Cprime%7D%20L%7D%7BA%7D)
![\frac{1}{2} \times \frac{(0.04)^{r} \times 4}{ L/{151}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B%280.04%29%5E%7Br%7D%20%5Ctimes%204%7D%7B%20L%2F%7B151%7D%7D)
solving above equation we get, elevation distance between highest point and pvc as 0.1208 ft
Answer:
When the brakes are applied the in the typical double transverse wishbone front suspension, it "drives" the car ground due to the setting of the link-type system pivot points on the lower wishbone are have parallel alignment to the road
Explanation:
In order to minimize the car's reaction to the application of the brakes, the front and rear pivot are arranged with the lower wishbone's rear pivot made to be higher than the front pivot as such the inclined wishbone torque results in an opposing vertical force to the transferred extra weight from the back due to breaking.
Engineering, maths and computer courses and more depending on your deficiencies