Hotter ocean tempatures mean more moisture in the dense air mass
Answer:
a) P =392.4[Pa]; b) F = 706.32[N]
Explanation:
With the input data of the problem we can calculate the area of the tank base
L = length = 10[m]
W = width = 18[cm] = 0.18[m]
A = W * L = 0.18*10
A = 1.8[m^2]
a)
Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:
P = density * g *h
where:
density = 1000[kg/m^3]
g = gravity = 9.81[m/s^2]
h = heigth = 4[cm] = 0.04[m]
P = 1000*9.81*0.04
P = 392.4[Pa]
The force can be easily calculated knowing the relationship between pressure and force:
P = F/A
F = P*A
F = 392.4*1.8
F = 706.32[N]
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
