The answer is Monocline. And I checked it, it's correct.
Given that,
Initial velocity , Vi = 0
Final velocity , Vf = 40 m/s
Acceleration due to gravity , a = 9.81 m/s²
Distance can be calculated as,
2as = Vf² - Vi²
2 * 9.81 *s = 40² - 0²
s = 81.55 m
For half height, that is, s = 40.77m
Vf= ??
2as = Vf² - Vi²
2 * 9.81 * 40.77 = Vf² - 0²
Vf² = 800
Vf = 28.28 m/s
Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.
Answer:
v = 12.4 [m/s]
Explanation:
With the speed and Area information, we can determine the volumetric flow.
where:
r = radius = 0.0120 [m]
v = 2.88 [m/s]
![A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2A%280.0120%29%5E%7B2%7D%20%5C%5CA%3D4.523%2A10%5E%7B-4%7D%20%5Bm%5D%5C%5C)
Therefore the flow is:
![V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]](https://tex.z-dn.net/?f=V%3D2.88%2A4.523%2A10%5E%7B-4%7D%20%5C%5CV%3D1.302%2A10%5E%7B-3%7D%20%5Bm%5E%7B3%7D%2Fs%20%5D)
Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.
![v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]](https://tex.z-dn.net/?f=v%3DV%2FA%5C%5Cv%3D1.302%2A10%5E%7B-3%7D%20%2F1.05%2A10%5E%7B-4%7D%20%5C%5Cv%3D12.4%5Bm%2Fs%5D)
Answer:
2.9 M
Explanation:
The concentration-time equation for a second order reaction is:
1/[A] = kt + 1/[A°]
Where,
A = concentration remaining at time, t
A° = initial concentration
k = rate constant
1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81
1/[A] = 0.345
= 1/0.345
= 2.9 M.
