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Olenka [21]
2 years ago
13

A furniture manufacturer purchases a drill press machine enabled with 5G and edge computing capabilities to keep the machine ope

rators safe when at work.How could these capabilities help improve safety of the operators?
Engineering
1 answer:
Andreas93 [3]2 years ago
8 0

A lot of manufacturer often uses 5G machines. How these capabilities could help improve safety of the operators is that it does includes an emergency switch for the operator so that one can manually shut off when needed.

<h3>Edge computing with 5G</h3>

  • The edge computing along with 5G network and IoT devices can help put together  different safety features and limitations and on can use them to known the unsafe action and also data can be communicated.

Edge computing when use with 5G produces good opportunities in all industry. It is known to help bring computation and data storage close to where data is been produced and it enable good data control, reduced costs, etc.

Learn more about 5G network from

brainly.com/question/24664177

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(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinuso
sammy [17]
Njjhvgghjjjhhhhhhhb hhh. I h. I’ve. Know
5 0
3 years ago
Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca
Andru [333]

Answer:

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

Explanation:

<u>a)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in series  

<u>List the Knowns: </u>

Capacitance of the first capacitor: C_{1}= 2цF = 2 x 10-6 F

Capacitance of the second capacitor C_{2}= 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

<u>Set Up the Problem:   </u>

Capacitance of a series combination:  

\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }+............

\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\

Capacitance of a series combination is given by:

C_{s}=\frac{Q}{V}

Then the charge stored in the series combination is:  

Q=C_{s} V

Energy stored in the series combination is:  

U_{c}=\frac{1}{2}  V^{2} C_{s}

<u>Solve the Problem:  </u>

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

<u>b)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in parallel  

<u>Set Up the Problem:  </u>

Capacitance of a parallel combination:

C_{p} =C_{1} +C_{2} +C_{3}

C_{p} =2+7.4=9.4*10^-6F

Capacitance of a parallel combination is given by

C_{p} =\frac{Q}{V}

Then the charge stored in the parallel combination is

Q=C_{p} V

Energy stored in the parallel combination is:  

U_{c}=\frac{1}{2} V^2C_{p}

<u>Solve the Problem: </u><em>  </em>

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

5 0
3 years ago
Read 2 more answers
A manufacturer makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't
aleksklad [387]

Answer:

probability P = 0.32

Explanation:

this is incomplete question

i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?

solution

we know that Standard error is expess as

Standard error = \frac{sd}{\sqrt{n}}

Standard error  = \frac{12}{\sqrt{16}}

Standard error  = 3  

so here we get Z value for 3 units away are from mean are

mean =  -1 and + 1

so here

probability P will be

probability P = P( z < -1 or z > 1)

probability P = 0.1587 + 0.1587

probability P =  0.3174

probability P = 0.32

7 0
4 years ago
Air pressure is higher above an airfoil.<br> true or false
attashe74 [19]

Answer: true

Explanation:

it flows faster over the top of the wing because the top is more curved than the bottom of the wing. However

6 0
3 years ago
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4    ( negative sign phase inversion)

b)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

so

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000  ( negative sign phase inversion)

3 0
3 years ago
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