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Nadya [2.5K]
3 years ago
11

What energy does a curcuit board run on

Engineering
2 answers:
Citrus2011 [14]3 years ago
8 0

Answer:

energy from computer

Explanation:

son4ous [18]3 years ago
6 0

a curcuit board is powered by energy from the computers power soarce

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What should you, the worker, be aware of with regard to evacuation procedures at your workplace
Alinara [238K]

Answer:

As a worker, it is important to follow the proper set of instructions or emergency plans during an emergent situation. Not carefully following the rules may result to a bigger problem such as further injury and damage to property.

Explanation:

Evacuation Procedure- This is a step-by-step procedure that people follow in order to safely vacate any building or place. This procedure is applicable to any situation, such as the workplace. This is now called the <em>Workplace Evacuation Procedure. </em>This is very important because there are so many unpredictable situations or events that are happening in the world right now, such as fire or earthquake. This procedure is being done through an evacuation plan.

The awareness of the workers regarding the proper way to evacuate during emergency situation is very important. It will be easier for them to know where to locate the nearest exit route. They will also learn to stop any form of device or equipment that could cause a hazzard during the situation. In case of the hospital, which is also a workplace, the employees will also learn how to assist the patients before themselves. They will also know where to assemble if there's a need to do so.

7 0
4 years ago
Read 2 more answers
The uniform slender rod has a mass m.
Nikolay [14]

Answer: 3/2mg

Explanation:

Express the moment equation about point B

MB = (M K)B

-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α

α = 3g/2L cosθ

express the force equation along n and t axes.

Ft = m (aG)t

mg cosθ – Bt = m [(3g/2L cos) (L/6)]

Bt = ¾ mg cosθ

Fn = m (aG)n

Bn -mgsinθ = m[ω^2 (L/6)]

Bn =1/6 mω^2 L + mgsinθ

Calculate the angular velocity of the rod

ω = √(3g/L sinθ)

when θ = 90°, calculate the values of Bt and Bn

Bt =3/4 mg cos90°

= 0

Bn =1/6m (3g/L)(L) + mg sin (9o°)

= 3/2mg

Hence, the reactive force at A is,

FA = √(02 +(3/2mg)^2

= 3/2 mg

The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg

6 0
3 years ago
An engineer is applying dimensional analysis to study the flow of air through this horizontal sudden contraction for the purpose
navik [9.2K]

Answer:

The answer is " \Delta p = f(V1, p, V2, d, D, L)"

Explanation:

Please find the complete question in the attached file.

Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.

The temperature change thus relies on V1 v2 p d D L.

4 0
3 years ago
Which contribution of Science is the most important discovery of man? Explain why?​
Vsevolod [243]

Answer:

<h2>The Invention of the Internal Combustion Engine (ICE)</h2>

Explanation:

The internal combustion engine is an engine in which ignition and combustion take place in the engine(in one place), the invention of the ICE was an integral part of the industrial revolution, as there was increasing demand for power, and manual labor could not suffice, especially during the mid 19 century.

The ICE made it possible for tasks that demand intensive power consumption to come through to reality, it was as a result of the invention of the ICE that road transportation was made easier for mankind, as the means of transport then was the use of beast of burden, now we have cars, airplanes ship, etc, essentially the invention of ICE reduced the tedious task man would have to engage in for his daily needs

5 0
3 years ago
Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan
melamori03 [73]

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

7 0
3 years ago
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