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Novosadov [1.4K]
3 years ago
13

In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

eights are connected by a thin 0.96-m rod with a mass of 24 kg. Find the moment of inertia of the barbell through the axis perpendicular to the rod at its center, assuming the two weights are small enough to be treated as point masses.
Physics
1 answer:
kondaur [170]3 years ago
6 0

The moment of inertia is 24.8 kg m^2

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

I_r = \frac{1}{12}ML^2

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2

The moment of inertia of one ball is given by

I_b = mr^2

where

m = 50 kg is the mass of the ball

r=\frac{L}{2}=\frac{0.96}{2}=0.48 m is the distance of each ball from the axis of rotation

So we have

I_b = (50)(0.48)^2=11.5 kg m^2

Therefore, the total moment of inertia of the system is

I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

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Answer:

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Explanation:

I hope this helps a little! :)

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The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo
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Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

W=\vec F\ \vec r

It can be written in its scalar version as

W=F.d.cos\theta

Being F and d the magnitudes of the force and displacement, and \theta the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

W=N.d.cos90^o=0

The work is zero because the force and the displacement are perpendicular

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Since in this case, there is no displacement, d=0, and the work is

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3 0
3 years ago
A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin
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Answer:

Explanation:

Given

mass of boy m_b=70.9\ kg

mass of girl m_g=43.2\ kg

speed of girl after push v_g=4.64\ m/s

Suppose speed of boy after push is v_b

initially momentum of system is zero so final momentum is also zero because momentum is conserved

P_i=P=f

0=m_b\cdot v_b+m_g\cdot v_g

v_b=-\frac{m_g}{m_b}\times v_g

v_b=-\frac{43.2}{70.9}\times 4.64  

v_b=-2.82\ m/s

i.e. velocity of boy is 2.82 m/s towards west                

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3 years ago
A reaction occurs in which carbon combines with oxygen to form carbon dioxide. Is this a chemical reaction or a nuclear reaction
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Answer:

A. This is a chemical reaction because only the electrons were rearranged.

Explanation:

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4 0
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The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

6 0
3 years ago
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