From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

m = mass

$v_{f,i}$ = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

$\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2$

$\frac{1}{2} (v_f^2-v_0^2) = gh_2$

$(v_f^2-v_0^2) = 2gh_2$

$v_f = \sqrt{2gh_2+v_0^2}$

$v_f = \sqrt{2(9.8)(23.5)+13.6^2}$

$v_f = 25.4m/s$

7 0

2 years ago

Apply the concept of density to explain why oil floats on water.

Sauron [17]

Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.

6 0

2 years ago