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2 years ago

14

A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a posi

tion with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na

1 answer:

ValentinkaMS [17]2 years ago

6 0

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3 x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 / 3 x .25

= 52.26

ω = 7.23 rad / s .

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In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a

xz_007 [3.2K]

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The correct answer is option a.

Explanation:

Conservation of momentum :

$m_1u_1+m_2u_2=m_1v_1+m_1v_2$

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$u_1,u_2$ = initial velocity before collision

$v_1,v_2$ = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

$m_1=m_2=M$

Initial velocity of $m_1=u_1=2 m/s$

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Final velocity of $m_1=v_1=v$ (same direction )

Final velocity of $m_2=v_2=v$ (same direction)

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The speed of the carts after their collision is 0.5 m/s.

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3 years ago

A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t

Mandarinka [93]

Answer:

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Explanation:

The Magnetomotive force (mmf) is given by the formula below

$F_{M} = Hl\\$

where $F_{M}$ is the Magnetomotive force (mmf)

$H$ is the Magnetic field strength

$l$ is the magnetic length

The magnetic permeability μ is given by

μ = $B / H$

Where $B$ is the Magnetic flux density

and $H$ is the Magnetic field strength

From the question,

$B$ = 1.2Wb/m^2

μ = 1600m

From μ = $B / H$

∴ $H = B/$ μ

$H = 1.2 / 1600\\$

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Now, for the Magnetomotive force (mmf)

$F_{M} = Hl\\$

From the question

$l$ = 1.5 m

∴ $F_{M} = 7.5$ × $10^{-4}$ × $1.5$

$F_{M} = 1.125$ × $10^{-3} A$

Hence, The mmf required is $1.125$ × $10^{-3}$ A

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3 years ago