Hi, it’s the iris. Hope this helped u
Answer:
K = ρL²g
Explanation:
Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;
Then:
F = upthrust ( restoring force) = weight of the liquid displaced.
![F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g](https://tex.z-dn.net/?f=F%20%3D%20V_%7B%5Comega%7D%20%5Crho_%7B%5Comega%7D%20g%3D%20A%20y%20%5Crho_%7B%5Comega%7D%20g)
where;
A = L²
![\rho_{\omega} = \rho](https://tex.z-dn.net/?f=%5Crho_%7B%5Comega%7D%20%3D%20%5Crho)
F = ky.
Then,
![Ay \rho g = ky](https://tex.z-dn.net/?f=Ay%20%5Crho%20g%20%3D%20ky)
![L^2y \rho g = ky](https://tex.z-dn.net/?f=L%5E2y%20%5Crho%20g%20%3D%20ky)
Divide both sides by y
K = ρL²g
Answer: 11 km/h at 339° compass
Explanation:
A sees B moving south at 0 km/h
A is moving north at 12cos30 = 10.392 km/h
Therefore B must be moving north at 10.392 k/h
A is moving east at 12sin30 = 6 km/h
B appears to be moving west at 10 km/h
Therefore B must be moving west at 10 - 6 = 4 km/h
B is moving v = √(4² + 10.392²) = 11.135... 11 km/h
θ = arctan( -4 / 10.392) = -21.05 = 339°
Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units