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makkiz [27]
3 years ago
9

The _______includes classes of laborers, such as general laborer, apprentice laborer and mason.

Engineering
1 answer:
____ [38]3 years ago
4 0

Answer:

Explanation:

lol why’d u put a blank we can read the missing word

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State 2 reasons on why blind spot checks are important
Marizza181 [45]

Answer:

How to Test for a Blind Spot . As you have learned, the blind spot is an area on your retina that has no visual receptors. Because of this, there is a tiny gap in your visual field. While your brain usually fills in the missing information so that you don't notice it, this quick and easy test makes it possible to demonstrate the blind spot , Of this, 300 crashes are, sadly, fatal. 2 Accidents are just as likely on Australian roads if blind spots are ignored. Not every car has blind-spot detection, so it's best to educate drivers about the importance of the blind spot to avoid accidents. However, not acknowledging the blind spot when driving isn't just dangerous for other cars on.

Explanation:

8 0
3 years ago
A machine raises 20kg of water through a height of 50m in 10secs. What is the power of the machine.​
Tomtit [17]

Answer:

hhahhhwghwhwhwhwjwnwjnnnnwnwwnw

Explanation:

jwkwkkwoiwiwiwiwiwowwiwowowiiiiwuuwuwgeevehehsvhsvwhbhhehehwgjjwhwhjwjqwjjuuuwi####!\\\\e

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3 years ago
Vital role of maritime english among seaferers
seropon [69]

Answer:

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8 0
3 years ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
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