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Ainat [17]
3 years ago
11

A small fan has blades that have a radius of 0.0600 m. When the fan is turned on, the tips of the blades have a tangential accel

eration of 23.7 m/s2 as the fan comes up to speed. What is the angular acceleration α of the blades?
Physics
1 answer:
zhuklara [117]3 years ago
7 0

Answer:

α = 395 rad/s²

Explanation:

Main features of uniformly accelerated circular motion

A body performs a uniformly accelerated circular motion   when its trajectory is a circle and its angular acceleration is constant  (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.

There is tangential acceleration (at) and is constant.

at = α*R     Formula (1)

where

α  is the angular acceleration

R is the radius of the circular path

There is normal or centripetal acceleration that determines the change in direction of the velocity vector.

Data

R = 0.0600 m   :blade radius

at = 23.7 m/s² : tangential acceleration of the blades

Angular acceleration of the blades (α)

We replace data in the formula (1)

at = α*R  

23.7 = α*(0.06)

α = (23.7) / (0.06)

α = 395 rad/s²

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3 years ago
A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
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Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

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*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

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