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Stolb23 [73]
1 year ago
5

characterizing vibration-assisted atomic force microscopy (afm)-based nanomachining via perception of acoustic emission phenomen

a using a sensor-based real-time monitoring approach
Physics
1 answer:
DaniilM [7]1 year ago
8 0

The study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

<h3>What is nanomachining?</h3>

The expression nanomachining makes reference to the study of nanometric machines (nanomachines) and related materials, which can be achieved by different approaches including sensor-based strategies related to acoustic auditive phenomena.

In conclusion, the study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

Learn more about nanomachines here:

brainly.com/question/20875598

#SPJ1

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A motorcycle starts to move from rest. If the velocity of the motorcycle becomes 90 km/hr
bagirrra123 [75]

Explanation:

90 kmhr—1 x 1000/3600 = 25ms—1

U = 0 ms—1

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2 years ago
A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0
Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

Explanation:

Given that,

Electric field E=-6.00\times10^{5}i\ N/C

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

F=F_{e}

ma=qE

a=\dfrac{Eq}{m}

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}

a=-5.74\times10^{13}\ m/s62

The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). We need to calculate the initial peed

Using equation of motion

v^2-u^2=2as

Where, s = distance

Put the value into the formula

0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}

u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}

u=2834783.94

u=2.83\times10^{6}\ m/s

The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

t=\dfrac{u}{a}

Where, u = initial velocity

a = acceleration

Put the value into the formula

t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}

t=0.493\times10^{-7}\ sec

Hence, (a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

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Nuclear power plants are fueled by uranium, which emits radioactive substances. Most of these substances are trapped in uranium fuel pellets or in sealed metal fuel rods. However, small amounts of these radioactive substances (mostly gases) become mixed with the water that is used to cool the reactor.

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