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____ [38]
3 years ago
9

In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, a

nd its applicability. It is recommended that you read the corresponding sections in your textbook before attempting these questionsn which of the following does the moment of inertia of an object depend? Check all that apply. linear speed linear acceleration angular speed angular acceleration total mass shape and density of the object location of the axis of rotation. b) What is the moment of inertia of particle a? mr2 9mr2 10mr2 undefined: an axis of rotation has not been specified.
Physics
1 answer:
scoundrel [369]3 years ago
4 0

Answer:

a)  Total mass form, density and axis of rotation location are  True

b)   I = m r²

Explanation:

a) The moment of inertia is the inertia of the rotational movement is defined as

       I = ∫ r² dm

Where r is the distance from the pivot point and m the difference in body mass

In general, mass is expressed through density

        ρ = m / V

        dm = ρ dV

From these two equations we can see that the moment of inertia depends on mass, density and distance

Let's examine the statements, the moment of inertia depends on

- Linear speed       False

- Acceleration angular False

-  Total mass form True

-  density True

- axis of rotation location   True

b) we calculate the moment of inertia of a particle

For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is

          I = m r²

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If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo
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A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
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Answer:

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Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

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The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

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-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

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3 years ago
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