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tino4ka555 [31]
3 years ago
9

A flat disk, a solid sphere, and a hollow sphere each have the same mass m and radius r. The three objects are arranged so that

an axis of rotation passes through the center of each object. The rotation axis is perpendicular to the plane of the flat disk. Which of the three objects has the largest moment of inertia?a. The solid sphere and hollow sphere are tied for largest moment of inertia.
b. The hollow sphere has the largest moment of inertia.
c. The solid sphere has the largest moment of inertia.
d. The flat disk has the largest moment of inertia.
e. The flat disk and hollow sphere are tied for largest moment of inertia.
Physics
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer:

Explanation:The rotational inertia of any object depends directly on the distance the mass is from the axis of a rotating object

Having more mass at the sides will increase the rotational inertia of the object that is why a Hollow sphere having same M and R as the solid one has more rotational inertia as it has more mass at the sides.

The  sphere have some mass at the center but most of its mass is closer to its radius and thus have more inertia than flat Disk.

The same relation exist between a flat disk and hollow sphere. The hollow sphere have some mass at the center but most of its mass is closer to its radius and thus have more inertia.

The rotational of the objects can be calculated by these equations

FlatDisk=\frac{1}{4} MR^2\\Sphere=\frac{2}{5} MR^2\\HollowSphere=\frac{2}{3} MR^2\\

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Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

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T = 12000K

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       P₂ / P₁ = T₂⁴ / T₁⁴

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Let's calculate the maximum emission wavelength

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T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

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An initial condition has been supplied by the problem: 

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<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

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<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

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0=(v)+(a)(t) 

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<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

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This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

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