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WITCHER [35]
4 years ago
12

As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94

rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Physics
1 answer:
lapo4ka [179]4 years ago
7 0

Answer:

I₂/I₁ = 0.53

Explanation:

During the motion the angular momentum of the skater remains conserved. Therefore:

Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms

L₁ = L₂

but, the formula for angular momentum is:

L = Iω

Therefore,

I₁ω₁ = I₂ω₂

I₂/I₁ = ω₁/ω₂

where,

I₁ = Initial Moment of Inertia

I₂ = Final Moment of Inertia

ω₁ = Initial Angular Velocity = 3.14 rad/s

ω₂ = Final Angular velocity = 5.94 rad/s

Therefore,

I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)

<u>I₂/I₁ = 0.53</u>

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Answer:

   q = 224 mm,   h ’= - 98 mm, real imagen

Explanation:

For this exercise let's use the constructor equation

        \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

       

where f is the focal length, p and q are the distance to the object and the image respectively.

In a mirror the focal length is

        f = R / 2

indicate us radius of curvature is equal to the diameter of the eye

       R = 3,50  10² mm

       f = 3.50 10² /2 = 1.75 10² mm

they also say that the distance to the object is p = 0.800 10³ mm

        1 / q = 1 / f - 1 / p

        1 / q = 1 / 175 - 1 /800

        1 / q = 0.004464

         q = 224 mm

to calculate the size let's use the magnification ratio

          m = \frac{h'}{h} = - \frac{q}{p}

          h '= - \frac{q}{p} \ h

          h ’= - 224 350 / 800

          h ’= - 98 mm

in concave mirrors the image is real.

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