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ladessa [460]
3 years ago
13

Calculate the speed (in m/s) of an electron and a proton with a kinetic energy of 1.85 electron volt (eV). (The electron and pro

ton masses are me = 9.11 ✕ 10−31 kg and mp = 1.67 ✕ 10−27 kg. Boltzmann's constant is kB = 1.38 ✕ 10−23 J/K.)
Physics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

ve = 8.06 x 10^5 m/s

vp = 1.882 x 10^4 m/s

Explanation:

K = 1.85 eV = 1.85 x 1.6 x 10^-19 J = 2.96 x 10^-19 J

me = 9.11 ✕ 10^-31 kg,

mp = 1.67 ✕ 10^-27 kg

Let the speed of electron is ve and the proton is vp.

For electron :

K = 1/2 me x ve^2

2.96 x 10^-19 = 0.5 x 9.11 x 10^-31 x ve^2

ve^2 = 6.498 x 10^11

ve = 8.06 x 10^5 m/s

For proton:

K = 1/2 mp x vp^2

2.96 x 10^-19 = 0.5 x 1.67 x 10^-27 x vp^2

vp^2 = 3.5449 x 10^8

vp = 1.882 x 10^4 m/s

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Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

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Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

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

70.632 - Mf = 0.4726•I + 2.0448

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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

1.0406•I + Mf = 132.274 equation 2

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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