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galina1969 [7]
3 years ago
15

Acquisition of resources from an external source is called?

Engineering
1 answer:
densk [106]3 years ago
6 0

Answer:

subcontracting

Explanation:

I hope this is right

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We know that passengers can be either helpful or harmful to a driver. Describe a pro and a con of having passengers in your car.
anygoal [31]

Answer:

sub to pewdipie

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3 years ago
Read 2 more answers
4. Which of the following are used for vehicle and/or major component identification?
AleksAgata [21]

Option C

VIN (vehicle identification number) are used for vehicle and/or major component identification

<h3><u>Explanation:</u></h3>

American automobile producers have practiced a vehicle identification number (V.I.N.) to name and recognize motor vehicles. They serve to retain an indication of obstacles, ownership exchanges, and prevent crime. This practice also posted the superimposed major component parts and replacement parts to be noted for each of the classes of vehicles.

Depending on the transport major parts may be marked with the vehicle’s VIN. The component part labels are composed of the alike material as the Federal Safety Certification Labels and are also planned to self destruct if excluded.

4 0
4 years ago
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure   to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = -\frac{dp}{dV/V}  ................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}   ............2

here ρ is density

and we know ρ  for 20°C = 998 kg/m³

and ρ  for 80°C = 972 kg/m³

so from equation 2 put all value

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

so now  % change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = \frac{\pi }{4} *d^2*l(i)    ................3

final volume v2 = \frac{\pi }{4} *d^2*l(f)    ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

put here all value

0.0130 = \frac{\pi }{4} *2^2*dl

dl = 0.004138 m

so water level rise is 4.138 mm

8 0
3 years ago
Someone who has the knowledge and experience to handle problems is a
goldfiish [28.3K]
Answer: Qualified person
If not give more context about the subject.
3 0
3 years ago
A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al
marysya [2.9K]

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

8 0
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