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kvasek [131]
3 years ago
5

Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge

, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed. In this final state
Physics
1 answer:
tatuchka [14]3 years ago
5 0

Answer:

The question is incomplete as some details are missing, here is the details ;

In this final state

a) spheres #1 and #2 both carry negative charge.

b) sphere #1 carries negative charge and #2 carries positive charge.

c) spheres #1 and #2 are still uncharged.

d) sphere #1 carries positive charge and #2 carries negative charge.

e) spheres #1 and #2 both carry positive charge.

Explanation:

From the concept of electrostatics, if a positively charged sphere is brought close to #2, there will be attraction of the opposite charges(-ve) towards it.

Now connecting a copper wire between #1 and #2, opposite charges will flow from #1 towards #2. disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charge and from the laws of attraction ; Like charges repel and unlike charges attract. the correct option is d.

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Suppose students experiment with the tube and a variety of darts. Some darts have higher masses than others but are the same aer
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The dart with the small mass will travel the farthest distance.

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In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
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Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

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When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

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R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

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Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

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a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

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M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

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M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

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0.4726•I + Mf = 70.632-2.0448

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Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

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Re arrange

1.0406•I + Mf = 141.264 - 8.99

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Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

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