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Rzqust [24]
3 years ago
11

The drawing shows three objects rotating about a vertical axis. The mass of each object is given in terms of m0, and its perpend

icular distance from the axis is specified in terms of r0. Rank the three objects according to their moments of inertia, largest to smallest.

Physics
1 answer:
photoshop1234 [79]3 years ago
7 0

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

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3 years ago
How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te
Tresset [83]

Answer:

<em>0.97c</em>

<em></em>

Explanation:

From the relativistic equation for length contraction, we have

l = l_{0}\sqrt{1 - \beta }

where

l is the final length of the object

l_{0} is the original length of the object before contraction

β = v^{2} /c^2

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

l/l_{0} = \sqrt{1 - \beta }

For the length to contract to one-fourth of the proper length, then

l/l_{0} = 1/4

substituting into the equation, we'll have

1/4 = \sqrt{1 - \beta }

substituting for β, we'll have

1/4 = \sqrt{1 - v^2/c^2 }

squaring both side of the equation, we'll have

1/16 = 1 - v^2/c^2

v^2/c^2 = 1 - 1/16

v^2/c^2 = 15/16

square root both sides of the equation, we have

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v = <em>0.97c</em>

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