V = [4/3]π r^3 => [dV / dr ] = 4π r^2
[dV/dt] = [dV/dr] * [dr/dt]
[dV/dt] = [4π r^2] * [ dr/ dt]
r = 60 mm, [dr / dt] = 4 mm/s
[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s
The solution for this problem is:
Remember that this doesn’t depend on the mass of the child.
E = T + U = constant
E (maximum height) = T + U =U = mgh = mg[r - r· cos (Θ)]
E (bottom height) = T + U = T = ½mv² = mg[r - r · cos (Θ)]
v² = 2g[r – r · cos (Θ)]
v = √ (2g[r-r·cos(Θ)])
= √(2(9.8)[3 – 3 · cos (45°)])
= 4.15 m/s or 15 kph
Answer:
P = 2161 N
Explanation:
For this exercise, let's start with Newton's second law
P - W = m a
P = ma + W
Where P is the fin force, W the weight,
Let's look for the vertical acceleration of the fish, this goes from a vertical speed of zero to a speed of 6.20 m / s when it has traveled half its length 1.50
y = 0.75 m
v² = v₀ + 2 a y
a = v² / 2 y
a = 6.20²/2 0.75
a = 25.63 m / s
We substitute in Newton's second law
P = m (g + a)
Let's calculate
P = 61.0 (0 9.8 + 25.63)
P = 2161 N