Calculate the approximate force on a square meter (1.00 m3) of sail, given the horizontal velocity of the wind is 6.00 m/s parallel to its front surface and 3.50 m/s along its back surface. Take the density of air to be 1.29 kg/m3. (The calculation, based on Bernoulli’s principle, is only approximate due to the effects of turbulence.) Remember, F = (P2 – P1)∙A

The force exerted on square meter of cubic sail is F = 15.3 N  

Explanation:

Given:-

- The face area of cubic sail, A = 1 m^2

- The velocity at frontal face, v1 = 6.0 m/s

- The velocity at back face, v2 = 3.5 m/s

- The density of air. ρ = 1.29 kg/m^3

Find:-

Calculate the approximate force on a square meter (1.00 m3) of sail

Solution:-

- We will apply the Bernoulli's equation to the flow of air around the cubic sail. Assuming that elevation changes are negligible. The constant elevation Bernoulli's equation is:

                           P1 + ρ*v1^2 / 2 = P2 + ρ*v2^2 / 2  

- The force (F) applied by any fluid is given by:

                           F = ( P2 - P1 )*A

- Re-arranging bernoulli's expression:

                          P2 - P1 = ρ/ 2 [ v2^2 - v1^2 ]

- Multiple the equation by area A:

                          A*[P2 - P1] = A*ρ/ 2 [ v1^2 - v2^2 ]

                          F = A*ρ/ 2 [ v1^2 - v2^2 ]

- Plug in the values:

                         F = (1)*(1.29/2)*[ 6^2 - 3.5^2 ]

                         F = 15.3 N  

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

where are the vectors ?