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2 years ago

7

The rotational inertia of an object depends not just upon the mass of the object but the mass distribution. What sort of mass di

stribution gives a large rotational inertia?

1 answer:

Oksanka [162]2 years ago

6 0

The object whose mass is distributed farther away from the axis of rotation will have greater rotational inertia.

<h3>What is rotational inertia?</h3>

Rotational inertia is a property of any object which can be rotated. It depends on mass and radius of the object.

<h3>Mass distribution of rotational inertia</h3>

The rotational inertia of two objects with the same mass and radius, is different when the mass distribution is different.

Thus, the object whose mass is distributed farther away from the axis of rotation will have greater rotational inertia.

Learn more about rotational inertia here: brainly.com/question/14001220

#SPJ1

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Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

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3 years ago

Which circuit component is usually a device that uses electrical energy to to perform a certain function? A. wire B. power sourc

Ksenya-84 [330]

Answer:

A

Explanation:

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3 years ago

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What single property was the most important in Jesseca’s material?

aleksandrvk [35]

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation:

Plato Answer

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3 years ago

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How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the $(velocity)^2$ and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

$\text { Kinetic Energy }=\frac{1}{2} m v^{2}$

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

$\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}$

$\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}$

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

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3 years ago

A cyclist going downhill is accelerating at 1.2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is t

Blababa [14]

Answer:

Initial Velocity is 4 m/s

Explanation:

What is acceleration?

It is the change in velocity with respect to time, or the rate of change of velocity.

We can write this as:

$a=\frac{\Delta v}{t}$

Where

a is the acceleration

v is velocity

t is time

$\Delta$ is "change in"

For this problem , we are given

a = 1.2

t = 10

Putting into formula, we get:

$a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12$

So, the change in velocity is 12 m/s

The change in velocity can also be written as:

$\Delta v = Final \ Velocity - Initial \ Velocity$

It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:

$\Delta v = Final \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4$

hence,

Initial Velocity is 4 m/s

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4 years ago

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