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3 years ago

High-mount stop lamps are typically attached to the exterior of the vehicle using:

Engineering

1 answer:

I am Lyosha [343]3 years ago

8 0

Answer:

Option A

Explanation:

Nylon retainers are knocking screws that have nylon coating on their thread which prevents them from loosening. It generally has two heads Hex Key an star screw. For high mount fixing, star screw is used as it is less prone to stripping. The nylon insert of a Nylock nuts enters into the screw and prevent it from loosening. It is useful in areas where there is high vibration.

Hence, option A is correct

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Define the coefficient of determination and discuss the impact you would expect it to have on your engineering decision-making b

scoundrel [369]

Answer and Explanation:

The coefficient of determination also called "goodness of fit" or R-squared(R²) is used in statistical measurements to understand the relationship between two variables such that changes in one variable affects the other. The level of relationship or the degree to which one affects the other is measured by 0 to 1 whereby 0 means no relationship at all and 1 means one totally affects the other while figures in between such 0.40 would mean one variable affects 40% of the other variable.

In making a decision as an engineer while using the coefficient of determination, one would try to understand the relationship between variables under consideration and make decisions based on figures obtained from calculating coefficient of determination. In other words when there is a 0 coefficient then there is no relationship between variables and an engineer would make his decisions with this in mind and vice versa.

7 0

3 years ago

Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure i

Brilliant_brown [7]

Answer:

$\dot Q_{in} = 372.239\,MW$

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

$w_{in} + h_{in}- h_{out} = 0$

$h_{out} = w_{in}+h_{in}$

$h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}$

$h_{out} = 203.81\,\frac{kJ}{kg}$

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

$h_{out} = 2685.4\,\frac{kJ}{kg}$

The rate of heat transfer in the boiler is:

$\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)$

$\dot Q_{in} = 372.239\,MW$

3 0

3 years ago

Read 2 more answers

What kinds of problems or projects would a civil engineer work on?

lisov135 [29]

Answer:

simple projects bovonhztisgx

8 0

3 years ago

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical

dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N

For Brass spacer

Pressure = 42201.6 N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 = 370.758 + 16^2

d^2 = 626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

5 0

3 years ago

A heat pump cycle whose coefficient of performance is 2.5 delivers energy by heat transfer to a dwelling at a rate of 20kW.

12345 [234]

Answer:

a) 8kW

b) $128

Explanation:

Given the coefficient of performance of the heat pump cycle to be 2.5

Energy delivered by the heat pump = 20kW

a) net power required to operate the heat pump = Energy delivered / coefficient of performance

Net power required = 20/2.5

= 8kW

b) Given the cost of electricity is $0.08 for 1kWhour

Since net power required to operate heat pump = 8kW

If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour

since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128

8 0

3 years ago