All categories

2 years ago

If we didn’t have the spark what could not happen?

Engineering

1 answer:

Sloan [31]2 years ago

8 0

Answer:

there's no electric current

You might be interested in

(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced

Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

See attached image file

5 0

2 years ago

3. How can statistical analysis of a dataset inform a design process?<br> PLEASE I NEED THIS ANSWER

julia-pushkina [17]

Answer:

you have to think then go scratch and then calculate and the design

Explanation:

4 0

2 years ago

An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens

Natalka [10]

Answer:

It will results in mechanical hardening.

5 0

3 years ago

Read 2 more answers

Your driver license will be _____ if you race another driver on a public road, commit a felony using a motor vehicle, or are fou

Georgia [21]

Hello there,

In the problems given in the question, the driver's license is confiscated and suspended.

So our answer is: A)

Achievements.

6 0

3 years ago

Read 2 more answers

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago