Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
Answer:
Element Is The Answer I think
C is the correct answer.
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Answer:
D) 15s
Explanation:
let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.
the period of a pendulum is given by:
T = 2π√(L/g)
so on earth:
Te = 2π√(L/g1)
= 6s
on the moon;
Tm = 2π√(L/g2)
since g2 = 1/6 g1 then:
Tm = 2π√(L/(1/6×g1))
= √(6)×2π√(L/(g1))
and 2π√(L/(g1)) = Te = 6s
Tm = (√(6))×6 = 14.7s ≈ 15s
Therefore, the period of the block-spring system on the moon is 15s.
