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I am Lyosha [343]
3 years ago
14

How many types of engineering specialist are there?

Engineering
1 answer:
klasskru [66]3 years ago
3 0

There are now six major branches of engineering: Mechanical, Chemical, Civil, Electrical, Management, and Geotechnical, and literally hundreds of different subcategories of engineering under each branch.

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A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
During the pre-drive check, you'll want to observe the car from the _______.
DENIUS [597]

Answer:

Passenger seat

Explanation:

If im wrong correct me

5 0
3 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he
inysia [295]

Given:

Temperature of water, T_{1} = -6^{\circ}C =273 +(-6) =267 K

Temperature surrounding refrigerator, T_{2} = 21^{\circ}C =273 + 21 =294 K

Specific heat given for water, C_{w} = 4.19 KJ/kg/K

Specific heat given for ice, C_{ice} = 2.1 KJ/kg/K

Latent heat of fusion,  L_{fusion} = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max COP_{refrigerator} = \frac{T_{2}}{T_{2} - T_{1}}

= \frac{267}{294 - 267} = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max COP_{heat pump} = \frac{T_{1}}{T_{2} - T_{1}}\frac{294}{294 - 267} = 10.89

6 0
3 years ago
You have a Switch LED Lightbulb with the following information: It is equivalent to a 40 Watt incandescent lightbulb. It uses 8
DochEvi [55]

The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.

<h3>How to calculate the lightning efficiency</h3>

The efficiency of the incandescent bulb will be:

= 450/40 = 11.25 lumens per watt.

The efficiency of the LED bulb will be:

= 450/8 = 56 lumens per watt.

In this case, the LED bulb is more efficient than the incandescent bulb.

Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient

Learn more lightning efficiency on:

brainly.com/question/25927632

4 0
2 years ago
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