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Step2247 [10]
3 years ago
12

Nikolai is using a hand-operated grain mill to grind wheatberries into flour. The mill is operated by spinning a fly-wheel with

radiusR= 23 cm, which has a handle attachedto the outer edge. After grinding for a few minutes at a con-stant angular speedωi, he lets go of the handle and allows themechanism to come to rest as it undergoes constant angularacceleration. This happens over the course oft= 0.50 s, andthe flywheel undergoes a quarter of a rotation during this time.What is the linear tangential accelerationaof the handle as itcomes to rest? For the limits check, investigate what happenstoaas the time required to stop the flywheel becomes small(t→0).
Physics
1 answer:
Harrizon [31]3 years ago
6 0

Answer:

Explanation:

α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²

ωf² = ωi² + 2αθ

θ = (ωf² - ωi²) / 2α

2π/4 = (0² - ωi²) / (2(-2ωi))

2π/4 = ωi / 4

ωi = 2π rad/s

α = -2(2π) = -4π rad/s²

a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²

as the time to stop approaches zero, acceleration goes toward infinity.

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Calculate the exerted on the heel of a boy's foot if the boy weighs 320N and he lands on one heel which has an area of 0.15m squ
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3 years ago
The horizontal surface on which the block of mass 2.2 kg slides is frictionless. The force of 27 N acts on the block in a horizo
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The magnitude of the resulting acceleration of the block is 6.14 m/s².

The given parameters;

  • <em>mass of the block, m = 2.2 kg</em>
  • <em>horizontal force on the block, F = 27 N</em>
  • <em>force below the horizontal, 81 N at 60⁰</em>

The net horizontal force on the block is calculated as follows;

F_x = 27 \ -  81 \times cos (60)\\\\F_x = -13.5 \ N

The acceleration of the block is calculated as follows;

a = \frac{-13.5}{2.2} \\\\a = - 6.14 \ m/s^2\\\\a = 6.14 \ m/s^2 \ to \ the \ left

Thus, the magnitude of the resulting acceleration of the block is 6.14 m/s².

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Which postulate of relativity was supported by the experiments of Michelson and Morley
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An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree
Mnenie [13.5K]

The y-component of the acceleration is 0.22 m/s^2

Explanation:

The y-component of the acceleration is given by

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u = 2.25 m/s is the initial velocity, in a direction \theta=50.0^{\circ}

v = 4.65 m/s is the final velocity, in a direction 120^{\circ}

t = 8.33 s is the time elapsed

The y-components of the initial and final velocity are:

u_y = u sin \theta = (2.25)(sin 50^{\circ})=1.72 m/s\\v_y = v sin \theta = (4.65)(sin 50^{\circ})=3.56 m/s

So the y-component of the acceleration is

a_y = \frac{3.56-1.72}{8.33}=0.22 m/s^2

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