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pashok25 [27]
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
Physics
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]

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

m_1u_1+m_2u_2=m_1v_1+m_1v_2

Where :

m_1, m_2 = masses of object collided

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

Initial velocity of m_2=u_2=-1 m/s (opposite direction)

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

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

M\times 2 m/s+M(-1 m/s)=Mv+Mv

1 m/s=2v

v = 0.5 m/s

rg135

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

7 0
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:

The mmf required is 1.125×10^{-3} A

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\\

H = 7.5 × 10^{-4}A/m

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

3 0
3 years ago
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