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A mass attached to a 66.4 cm long string starts from rest and is rotated 36.6 rev in1 min before reaching a final angular speed.
1 answer:
Answer:
0.128 rad/s², 7.66 rad/s
Explanation:
length, l = 66.4 cm
initial angular velocity, ωo = 0 rad/s
Let ω represent the final angular velocity.
Let α denote the angular acceleration.
number of revolutions, n = 36.6
time taken, t = 1 min = 60 seconds
Angle rotated, θ = 2πn = 2 x 3.14 x 36.6 = 229.85 rad
Apply the second equation of motion for angular dynamics
229.85 = 0 + 0.5 x α x 60 x 60
α = 0.128 rad/s²
Utilize the first equation of motion
ω = ωo + αt
ω = 0 + 0.128 x 60
ω = 7.66 rad/s
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Answer:
(a) the coefficient of friction is 0.451
This was derived using the energy conservation principle (the total energy in a closed system remains constant).
(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.
Explanation:
For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.
Thank you for your attention; I trust this is beneficial to you.
Details that are not provided: the problem figure is included.
We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,
where:
represents the pressure difference across the two ends
denotes the viscosity of the fluid
L signifies the length of the pipe
indicates the volumetric flow rate, where
is the cross-sectional area of the tube and
refers to the fluid's velocity
r stands for the pipe's radius.
This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:
is the dynamic viscosity of water at
L=4.0 cm=0.04 m
and r=1 mm=0.001 m
Substituting these values into the formula yields:
This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.
