Plz slv this answer

Fig.3 shows three 10 g particles that have been glued to a rod of length L = 6 cm and negligible mass. The assembly can rotate a

Phoenix [80]

(a) The amount of work required to change the rotational rate is 0.0112 J.

(b) The decrease in the rotational inertia when the outermost particle is removed is 64.29%.

<h3>Moment of inertia of the rod</h3>

The moment of inertia of the rod from the axis of rotation is calculated as follows;

I = md² + m(2d)² + m(3d)²

where;

m is mass = 10 g = 0.01 kg
d = 3 equal division of the length

d = 6/3 = 2 cm = 0.02 m

I = md²(1 + 2² + 3²)

I = 14md²

I = 14(0.01)(0.02)²

I = 5.6 x 10⁻⁵ kg/m³

<h3>Work done to change the rotational rate</h3>

K.E = ¹/₂Iω²

K.E = ¹/₂(5.6 x 10⁻⁵)(60 - 40)²

K.E = ¹/₂(5.6 x 10⁻⁵)(20)²

K.E = 0.0112 J

<h3>Percentage decrease of rotational inertia when the outermost particle is removed</h3>

I₂ = md² + m(2d)²

I₂ = 5md²

ΔI = 14md² - 5md²

ΔI = 9md²

η = (ΔI/I) x 100%

η = (9md²/14md²) x 100%

η = 64.29 %

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

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4 0

1 year ago

How does the junkyard magnet let go of metal?

klemol [59]

Answer:

Wrecking yards employ extremely powerful electromagnets to move heavy pieces of scrap metal or even entire cars from one place to another. ... This creates a magnetic field around the coiled wire, magnetizing the metal as if it were a permanent magnet

Explanation:

7 0

3 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$