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dalvyx [7]
2 years ago
10

Not sure which one....

Engineering
1 answer:
Airida [17]2 years ago
4 0
I think downwards as that's how most saw's work.
You might be interested in
1)What are the three previous manufacturing revolutions Mr. Scalabre mentions? When did these take place?
Ostrovityanka [42]

The three previous manufacturing revolutions that Mr. Scalabre mentioned and their year of occurrence are:

  1. The steam engine in the mid-19th Century
  2. The mass-production model in the early 20th Century
  3. The first automation wave in the 1970s

<h3>What is a Manufacturing Revolution?</h3>

This refers to the process of change from a handicraft economy to industry production-based production.

Hence, we can see that Mr. Scalabre believes we are not growing in productivity because there has not been enough automation to perform the tasks needed.

The effect of robotics is making an impact on productivity because a lot of complex, difficult tasks are done by machines.

3D printing has made an impact on productivity because there is a reduction in the pressing cycle and errors due to negligence are reduced.

The role the engineers have to play in the next revolution is that they would have to produce mathematical model that can be used to produce better AIs

Read more about manufacturing revolutions here:

brainly.com/question/14316656

#SPJ1

8 0
1 year ago
It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of
monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                         M_c = F_a*( 42 / 150 ) *88

                         M_c = 2.5*( 42 / 150 ) *88

                         M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0
3 years ago
Write a program to control the operation of the RED/GREEN/BLUE LED (LED2) as follows: 1. If no button is pressed, the LED should
aalyn [17]

Answer:

See explaination

Explanation:

int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }

4 0
3 years ago
A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Dy345-kV, Y transformer to a 345-kV circ
aniked [119]

Answer:

(a) the subtransient current through the breaker in per-unit and in kA rms =   71316.39kA

(b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. = 152KA

Explanation:

check the attached files for explanation

7 0
3 years ago
A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

7 0
3 years ago
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