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Kazeer [188]
2 years ago
14

True or False? Early engineers used a trial-and­error approach, rather than mathematical and scientific principles when solving

problems.
Engineering
1 answer:
yarga [219]2 years ago
3 0

Answer:

<h2>True Most Especially in the field of Automotive Engineering</h2>

Explanation:

Normally, before the introduction of vehicle diagnostics when a vehicle, mostly automobile/car break down, one could be the vehicle mechanic would only suspect one or two related faults based on the present working condition of the car, the mechanic would perform some trial and error before he could fix the car.

But in recent times, the introduction of vehicle diagnostics devices and software has changed the order as vehicles can be connected to a computer that will scan and tell what   the problem is before a possible fix.

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Lets Try This: study the pictures. Describe what you see and think about it. write your answer on a sheet of paper. home room
Yuliya22 [10]

Answer: I see three children cleaning the lawn while one of them are raking the leaves and one is holding a dust pan. The other child is holding a bucket. On the other picture, i see a young boy watering plants.

BTW: these pictures are not very clear so answers may vary.

Explanation:

4 0
2 years ago
Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.
Mekhanik [1.2K]

Answer:

98,614.82 W/m²

Explanation:

Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

Where;

Q = the amount of heat loss from the pipe

h =  the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂  = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})

\frac{Q}{L} = 314.159(\frac{70}{0.223})

\frac{Q}{L} = 314.159(313.901) = 98,614.82 W/m²

3 0
3 years ago
A 40kg steel casting (Cp=0.5kJkg-1K-1) at a temperature of 4500C is quenched in 150kg of oil (Cp=2.5kJkg-1K-1) at 250C. If there
podryga [215]
Of the oil I hope this help
4 0
3 years ago
The controller determines if a(n) _________ exists by calculating the difference between the SP and the PV.
shusha [124]

The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

<h3>How does a controller work in control system?</h3>

The Control system is one where it entails if the output is one that has an effect on the input quantity.

So it uses the PV(Process Variable) set against the SP(Setpoint) to know if an error exists.

So, The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

Learn more about controller from

brainly.com/question/14617664

#SPJ1

7 0
2 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
2 years ago
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