Use bernoulli's equation to find out how fast water leaves an opening in a water tank. the water level is 0.75 m above the opening. the top of the water tank is opened to the atmosphere.

3.84 m/s

Explanation:

Using Bernoulli's equation below:

P1 + (1/2ρv1²) + h1ρg = P2 + (1/2ρv2²) + h2ρg

where P1 = P2 atmospheric pressure

(1/2ρv1²) + h1ρg = (1/2ρv2²) + h2pg

collect the like terms

h1ρg - h2ρg = (1/2ρv2²) - (1/2ρv1²)

factorize the expression by removing the like terms on both sides

gρ(h1 - h2) = 1/2ρ( v2² - v1²)

divide both side by rho (density in kg/m³, ρ )

g(h1 - h2) = 1/2 (v2² - v1²)

assuming the surface of the tank is large and the speed of water then at the tank surface  v1 = 0

2g(h1 - h2) = v2²

take the square root of both side and h1 - h2 is the difference between the surface of the tank and the opening where water is coming out in meters

√2g(h1 - h2) = √ v2²

v2 = √2g(h1-h2) = √ 2 × 9.81×0.75 = 3.84 m/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards