Ask question

Ask question

All categories

LUCKY_DIMON [66]

3 years ago

10

A skilled worker with the ability to operate computer numerically controlled (CNC) machines is qualified to work in which of the

following jobs?
assembly worker
machinist
semi-conductor technician
engineer

1 answer:

KengaRu [80]3 years ago

3 0

Answer:

Machinist

Explanation:

A skilled worker with the ability to operate computer numerically controlled (CNC) machines is qualified to work in a machinist position.

A machinist is a person who is properly skilled and consists of advanced knowledge regarding the functions of a CNC machine. He can use different mechanisms and complex numerical functions of the machine to carry out different tasks. Any person who lacks the official learning of mechanisms cannot operate such machines effectively.

You might be interested in

In a short essay, discuss the question, "How are you an innovator?"

iragen [17]

Answer:

Being innovative means doing things differently or doing things that have never been done before. An innovator is someone who has embraced this idea and creates environments in which employees are given the tools and resources to challenge the status quo, push boundaries and achieve growth.

Explanation:

Hope it helps..

But it's a little bit long..

Correct me if I'm wrong..

7 0

3 years ago

1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur

Anit [1.1K]

Answer:

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, $T_L = 36^0 F = 275.372 K$

Outside temperature, $T_H = 100^0F = 310.928 K$

rate of insulation, $Q = 100 Btu/h$

To get the minimum electrical power required, use the relation below:

$\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt$

V = 5 V

Power = IV

$W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A$

If the cooler is supposed to work for 4 hours, t = 4 hours

$I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr$

Minimum battery size needed = 3.03 Amp-hr

6 0

3 years ago

Which starting circuit uses fuses, switches, and smaller wires to energize a relay and solenoid?

zysi [14]

Answer:

Option B (Starter Control Circuit) is the right option.

Explanation:

This same switching is normally put upon this isolated side of something like the transmission Arduino microcontroller throughout the configuration that is using the ignition just to command the broadcast.
It uses a secondary relay isolated to regulate electrical current throughout the solenoid starting system.

All other given options are not related to the given instance. So the above option is correct.

4 0

3 years ago

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and

belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = $\frac{wet mass - dry mass }{wet mass}$ = $\frac{0.525 - 0.49}{0.525}$ = 0.07 or 7%

water content = $\frac{wet mass - dry mass}{wet mass}$ = 7%

8 0

3 years ago

Read 2 more answers

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

3 0

3 years ago