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denpristay [2]
4 years ago
5

A wall clock has a minute hand with a length of 0.53 m and an hour hand with a length of 0.26 m. Take the center of the clock as

the origin, and use a Cartesian coordinate system with the positive x axis pointing to 3 o'clock and the positive y axis pointing to 12 o'clock. What is the magnitude of the acceleration of the tip of the minute hand of the clock?
Physics
1 answer:
ollegr [7]4 years ago
3 0

Answer:

a  =1.61 × 10⁻⁶ m/s²

Explanation:

given,

length of minute hand = 0.53 m

length of hour hand = 0.26 m

the time taken by the minute hand to complete one revolution is T=3600 s

the angular frequency is

                        \omega =\dfrac{2\pi }{T}

                         \omega =\dfrac{2\pi }{3600}

                         \omega=0.001745 rad/sec

the acceleration is  

                                a = r \omega^2

                                a = (0.53) \times (0.001745)^2

                                       a  =1.61 × 10⁻⁶ m/s²

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aliya0001 [1]

Answer:

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Explanation:

<h3>τ=Iα</h3><h3>τ= torque</h3><h3>I = inertia</h3><h3>α= angular acceleration</h3>

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when the pencil is released it will experience force due to weight

F=mgsinθ

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How can acceleration be involved if velocity is constant?
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alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

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As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of  this force, along these directions.

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While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

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This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which  is given by the following expression:

fsmax = μN (2)

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fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

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