a trained sea lion slides from rest down a long
ramp into a pool of water. if the ramp is inclined at
an angle of 23° above the horizontal and the
coefficient of kinetic friction between the sea lion
and ramp is 0.26. calculate the acceleration of the
sea lion as it slides down the ramp.

Given:

By using the Newton's second equation, we get

→

or,

→

→      

hence,

The acceleration will be:

→

Thus the above response is correct.

Learn more about newton's law here:

link

1.5 m/s²

Explanation:

Draw a free body diagram.  There are three forces acting on the sea lion: gravity pulling down, normal force perpendicular to the ramp, and friction parallel to the ramp.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substitute for N:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

a = g (sin θ − μ cos θ)

Given θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

a = 1.48

Rounded to two significant figures, the sea lion accelerates at 1.5 m/s².

i would think form in reponse to tensional forces

gas and liquid