Answer: 0.5 seconds or 2.625 seconds
Explanation:
At t = 0, The ball is 4 ft above the ground.
The height of the football varies with time in the following way:
s(t) = -16 t² + 50 t + 4
we need to find the time in which the height would of the football would be 25 ft:
⇒25 = -16 t² + 50 t + 4
we need to solve the quadratic equation:
⇒ 16 t² - 50 t + 21 = 0
⇒ t = 0.5 s or 2.625 s
Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.
Answer:
perpendicular to the direction of the greatest stress
Answer:
x = 4,138 m
Explanation:
For this exercise, let's use the rotational equilibrium equation.
Let's fix our frame of reference on the left side of the pivot, the positive direction for anti-clockwise rotation
∑ τ = 0
n₁ 0 - W L / 2 + n₂ 4 - W_woman x = 0
x = (- W L / 2 + 4n2) / W_woman
Let's reduce the magnitudes to the SI System
M = 6 lbs (1 kg / 2.2 lb) = 2.72 kg
M_woman = 130 lbs = 59.09 kg
Let's write the transnational equilibrium equation
n₁ + n₂ - W - W_woman = 0
n₁ + n₂ = W + W_woman
n₁ + n₂ = (2.72 + 59.09) 9.8
At the point where the system begins to rotate, pivot 1 has no force on it, so its relation must be zero (n₁ = 0)
n₂ = 605,738 N
Let's calculate
x = (-2.72 9.8 6/2 + 4 605.738) / 59.09 9.8
x = 4,138 m
Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
