Answer:
The gate will open if the height of water is equal to or more than 0.337m.
Explanation:
From the diagram attached, (as seen from the reference question found on google)
The forces are given as
Force on OA

Here
- ρ is the density of water.
- g is the gravitational acceleration constant
is the equivalent height given as

is the area of the OA part of the door which is calculated as follows:
The Force is given as
![F_1=0.6\rho g[h+0.3]](https://tex.z-dn.net/?f=F_1%3D0.6%5Crho%20g%5Bh%2B0.3%5D)
Force on OB

Here
- ρ is the density of water.
- g is the gravitational acceleration constant
is the equivalent height given as

is the area of the OB part of the door which is calculated as follows:

The Force is given as
![F_2=0.4\rho g[h+0.8]](https://tex.z-dn.net/?f=F_2%3D0.4%5Crho%20g%5Bh%2B0.8%5D)
Now the moment arms are given as
![\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}](https://tex.z-dn.net/?f=%5Cbar%7By%7D_a%3D%5Cbar%7Bh%7D%2B%5Cfrac%7BI%7D%7BA%5Cbar%7Bh%7D%7D%5C%5C%5Cbar%7By%7D_a%3Dh%2B0.3%2B%5Cfrac%7B%5Cfrac%7B1%7D%7B12%7D%5Ctimes%200.6%5E3%20%5Ctimes%201%7D%7B0.6%20%5Ctimes%5Bh%2B0.3%5D%7D%5C%5C%5Cbar%7By%7D_a%3Dh%2B0.3%2B%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D)
![\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}](https://tex.z-dn.net/?f=%5Cbar%7By%7D_b%3D%5Cbar%7Bh%7D%2B%5Cfrac%7BI%7D%7BA%5Cbar%7Bh%7D%7D%5C%5C%5Cbar%7By%7D_b%3Dh%2B0.8%2B%5Cfrac%7B%5Cfrac%7B1%7D%7B12%7D%5Ctimes%200.4%5E3%20%5Ctimes%201%7D%7B0.4%20%5Ctimes%5Bh%2B0.8%5D%7D%5C%5C%5Cbar%7By%7D_b%3Dh%2B0.8%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D)
Taking moment about the point O as zero
=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m](https://tex.z-dn.net/?f=F_1%28h%2B0.6-%5Cbar%7By%7D_a%29%3DF_2%28%5Cbar%7By%7D_b-h%2B0.6%29%5C%5CF_1%28h%2B0.6-h-0.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3DF_2%28h%2B0.8%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D-h-0.6%29%5C%5CF_1%280.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3DF_2%280.2%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D%29%5C%5C0.6%5Crho%20g%5Bh%2B0.3%5D%280.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3D0.4%5Crho%20g%5Bh%2B0.8%5D%280.2%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D%29%5C%5C0.6%5Bh%2B0.3%5D%280.3-%5Cfrac%7B0.03%7D%7Bh%2B0.3%7D%29%3D0.4%5Bh%2B0.8%5D%280.2%2B%5Cfrac%7B0.0133%7D%7Bh%2B0.8%7D%29%5C%5C0.18h%20-0.054-0.018%3D0.08h%2B0.064%2B0.00533%5C%5Ch%3D0.337%20m)
So the gate will open if the height of water is equal to or more than 0.337m.
Answer:
Fgparallel = 170N
Explanation:
Acting down the incline would be the paralell force. The Force of gravity on an incline for the parallel portion is mgsinθ.
Fg parallel = mgsinθ
mg is 289.9, as that is the weight. θ is 35.9
Fgpar = 289.9sin35.9
Fgparallel = 170N
Explanation:
The gravitational potential energy is given by :
P = mgh
The kinetic energy of an object is given by :

As the ball reaches the bottom of the ramp, its potential energy decreases and kinetic energy increases.
It imply that, when the ball at the top most height, its gravitational potential energy is maximum and zero kinetic energy and when ball reaches the bottom of the ramp, it will have maximum kinetic energy and zero potential energy.
Answer:
Graph C
Explanation:
This is the answer because it is the only one that shows the vehicle accelerate to a constant speed before stopping and slowing down.