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4 years ago

When encountering low visibility from rain or fog, you should use your ____.

Engineering

1 answer:

kirza4 [7]4 years ago

4 0

Answer:

Explanation:

Low beams should only be used when fog and rain is present, as high beams can cause a dangerous glare to you and other drivers. You should also use your fog lights, but not every vehicle has them.

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A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Po

kicyunya [14]

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

$E^z$ = - ( εˣ / v )

$E^z$ = - ( -0.00088 / 0.34 )

$E^z$ = - ( - 0.002588 )

$E^z$ = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

8 0

3 years ago

Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c

vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

$COP_{i} = \frac{250\,K}{300\,K-250\,K}$

$COP_{i} = 5$

The real Coefficient of Performance is:

$COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}$

$COP_{r} = 12.571$

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0

3 years ago

Who is the mystical body of Christ

Sergeu [11.5K]

If you're talking ab jesus christ , some religions believe it was god in human form and some believe it was gods son it all depends on the religion and how it's taught

8 0

3 years ago

Read 2 more answers

Soils with low percolation rates do not need special attention during site engineering. select one: true false

saveliy_v [14]

It is accurate to say that site engineering does not require particular consideration for soils with low percolation rates.

<h3>What are percolation rates?</h3>

The rate at which water percolates through the soil is a measure of its ability to absorb and treat effluent, or wastewater that has undergone preliminary treatment in a septic tank.
Minutes per inch are used to measure percolation rate (mpi).
The process of a liquid gently moving through a filter is called percolation. This is how coffee is typically brewed.

The Latin verb percolare, which meaning "to strain through," is the source of the word "percolation." When liquid is strained through a filter, such as when making coffee, percolation occurs.

To learn more about percolation rates, refer to:

brainly.com/question/28170860

#SPJ4

7 0

2 years ago

A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig

jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

Section: 9-ft section of pipe.

Sum of forces perpendicular to member AC = 0

F_d - 0.8*W*L = 0

F_d = 0.8*10*9 = 72 lb

Sum of forces perpendicular to member BC = 0

F_e - 0.6*W*L = 0

F_e = 0.6*10*9 = 54 lb

F_d = 72 lb , F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

Section: Entire Frame.

Sum of moments about point B = 0

-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

-A_y*(1.5625) + 15 + 39.375 = 0

A_y = 34.8 lb

Sum of forces in vertical direction = 0

A_y + B_y - 0.8*F_d - 0.6*F_e = 0

B_y = 0.8*(72) + 0.6*(54) - 34.8

B_y = 55.2 lb

Sum of forces in horizontal direction = 0

A_x + B_x - 0.6*F_d + 0.8*F_e = 0

A_x + B_x = 0

Section: Member AC

Sum of moments about point C = 0

F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

72*2.5 - 34.8*12 - 9*A_x = 0

A_x = -237.6 / 9 = - 26.4 lb

B_x = - A_x = 26.4 lb

A_x = -26.4 lb , B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

Member AC:

From point A till just before point D

-0.6*A_x*x - A_y*0.8*x + M = 0

15.84*x - 27.84*x + M = 0

M = 12*x ..... max value at D, x = 12.25 in

M_max = 12*12.25/12 = 12.25 lb-ft

Member BC:

From point B till just before point E

-0.8*B_x*x + B_y*0.6*x + M = 0

-21.12*x + 33.12*x + M = 0

M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in

M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

5 0

3 years ago