Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.
Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.
Answer:
a) Initial Value Problem
dv/dt = 4 - 0.1v
v(0) = 0
b) solution to the IVP
v(t) = 40(1 - e^(-t/10))
c) Limiting velocity
Vo = 40 ft/s
Position of the car after 12 hours
X = 14,390 ft
Explanations:
The complete explanations of each of the sections contained in the question are in the files attached to this solution.
Answer:
<em>Sonogram </em><em>is </em><em>a </em><em>medical </em><em>image </em><em>produced </em><em>by </em><em>ultrasound </em><em>echo. </em>
<em>It </em><em>is </em><em>used </em><em>to </em><em>help</em><em> </em><em>diagnose </em><em>causes </em><em>of </em><em>pain </em><em>and </em><em>swelling</em><em>. </em>
Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s
Let ω₂ = the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
= (0.7/3.5)*(30.159)
= 6.032 rad/s
ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s
Answer: 0.96 rev/s
Answer:
τsolid = 0.15 N•m
τhoop = 0.30 N•m
Explanation:
θ = ½αt²
α = 2θ/t² = 2(14)/8.3² = 0.406445 rad/s²
Solid disk I = ½mr² = ½(4.2)0.42² = 0.37044 kg•m²
τ = Iα = 0.37044(0.406445) = 0.150563... N•m
Hoop disk I = mr² = (4.2)0.42² = 0.74088 kg•m²
τ = Iα = 0.74088(0.406445) = 0.301127... N•m
