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1 year ago

Where can you go to find the system requirements?

Engineering

1 answer:

SSSSS [86.1K]1 year ago

8 0

Answer:c

Explanation:

yea, jus go to the systems help center then youll be confronted with a talks. person n they'll walk u threw it.

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9. Calculate the total resistance and current in a parallel cir-

Taya2010 [7]

Answer:

d. 2.3 ohms (5.3 amperes)

Explanation:

The calculator's 1/x key makes it convenient to calculate parallel resistance.

Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms

This corresponds to answer choice D.

<em>Additional comment</em>

This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.

5 0

3 years ago

Ughhh my cramps hurt sm

kozerog [31]

Answer:

Explanation:Come

tomate

5 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

4 years ago

Which type of engineer is needed in the following scenario?

gregori [183]

Answer:

A: Agricultural Engineer

Explanation:

I had this same question for a test and got it right with a being the answer :)

8 0

3 years ago

Read 2 more answers

Can you prove that the two bleu areas are the same without numbers please?

Svet_ta [14]

Answer:

$\small{\boxed{\tt{\colorbox{green}{✓Verified\:answer}}}}\:$

Just draw a line from point D join to point E

The triangle formed DME will be congruent to AMC

6 0

3 years ago