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Lerok [7]
3 years ago
9

What is the automated system that uses an automated work cell controlled by electronic signals from a common centralized compute

r facility?
Physics
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

The automated system that uses an automated work cell controlled by electronic signals from a common centralized computer facility is "Robotics"

Explanation:

Robotics is an advanced technology fully automated which uses electronic sensors incorporated with the combination of control into mechanical systems greatly enhancing the performance and flexibility of the systems. This is possible with the advances in hardware, software, and control programming systems which amounts to extensive automation from a common centralized computer facility.

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Ulleksa [173]

Answer:

12

Explanation:

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5 0
3 years ago
Read 2 more answers
A proton in a particle accelerator is traveling at a speed of 0.99c.(a) If you use the approximate nonrelativistic equation for
____ [38]

Answer:

a)  p = 4.96 10⁻¹⁹ kg m / s , b)  p = 35 .18 10⁻¹⁹  kg m / s ,

c)  p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

                 p = m v

Let's calculate its value

               p = 1.67 10⁻²⁷ 0.99 3. 10⁸

               p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

               p = m v / √(1 –v² / c²)

Let's calculate

                p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

                p = 4.96 10⁻¹⁹ / 0.141

                p = 35 .18 10⁻¹⁹  kg m / s

c) the relationship between the two values ​​is

            p_correst / p_approximate = 35.18 / 4.96

            p_correst / p_approximate = 7.09

4 0
3 years ago
Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
3 years ago
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

6 0
2 years ago
A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f
FinnZ [79.3K]

Answer:

{P_2}-P_1=49.99\ KPa

Explanation:

v_1=9.43\ m/s

d_1=11.7\ cm/s

d_2=17.5\ cm/s

From continuity equation

A_1v_1=A_2v_2

v_1d_1^2=v_2d_2^2

v_2=\dfrac{9.43\times 11.7^2}{17.5^2}

v_2=4.21\ m/s

\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2

{P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g

\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1

1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1

{P_2}-P_1=49.99\ KPa

4 0
3 years ago
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