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tino4ka555 [31]
2 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]2 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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When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f
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Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (r), measured in kilometers, can be represented by a right triangle:

r = \sqrt{x^{2}+y^{2}} (1)

Where:

x - Horizontal distance between the rocket and the observer, measured in kilometers.

y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}

\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}

\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}

Where:

\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

\dot y - Vertical speed of the rocket, measured in kilometers per hour.

If we know that y = 4\,km and \dot y = 400\,\frac{km}{h}, then the rate of change of the angle of elevation is:

\dot \theta = 48.780\,\frac{\circ}{s}

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0
3 years ago
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