A group of individuals living in a particular geographic area is termed population.
Answer:
The period of motion of new mass T = 0.637 sec
Explanation:
Given data
Mass of object (m) = 9 gm = 0.009 kg
Δx = 3.5 cm = 0.035 m
We know that spring force is given by
F = m g
F = 0.009 × 9.81 = 0.08829 N
Spring constant
k = 2.522 
New mass
= 26 gm = 0.026 kg
Now the period of motion is given by
T = 0.637 sec
This is the period of motion of new mass.
The formula we use
here is:
radial acceleration =
ω^2 * R <span>
110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>
<span>and since: ω = 2pi*f --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
<span>= 36,714.77 rpm </span></span>
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
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