Answer:
1.5 m/s²
Explanation:
Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.
Considering the forces perpendicular to the incline:
∑F = ma
N − mg cos θ = 0
This gives us N = mg cos θ
Next, examining the forces parallel to the incline:
∑F = ma
mg sin θ − Nμ = ma
Substituting for N yields:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g cos θ μ = a
hence a = g (sin θ − μ cos θ)
If we set θ = 23° and μ = 0.26:
a = 9.8 (sin 23 − 0.26 cos 23)
this results in a = 1.48
When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².
True. Explanation: In this instance, the area of the graph represents the impulse. Impulse is defined as the change in an object's momentum. Moreover, it is also expressed as the product of the force acting on an object and the duration of the impact. When we graph the force against time, if the force remains constant, the resultant graph will take on a rectangular shape, and the area under that graph will equal the impulse's definition.
Answer:
F = 0.535 N
Explanation:
We will apply energy concepts, considering both the peak and the bottom of the path.
Top
Em₀ = U = mg y
Bottom
= K = ½ m v²
Emo =
mg y = ½ m v²
v = √ (2gy)
y = L - L cos θ
v = √ (2g L (1 - cos θ))
Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.
F = ma
a = v² / r
For the turning radius, the cable length is r = L.
F = m 2g (1 - cos θ)
Now, let's find the result.
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
