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worty [1.4K]
3 years ago
7

In remote areas, your gps devices may lose reception. It’s a good idea to have a

Engineering
2 answers:
Katarina [22]3 years ago
4 0

Answer: MAP!!! it's always a good Idea to have a map nearby, cause you'll never know when you'll need it or like you said, lose reception

Serhud [2]3 years ago
3 0

Answer:

You will need a map, having a map is always very important when electronic gadgets don't work, but you can also take a compass with you.

Explanation:

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12. What procedure should you follow when taking measurements?
OlgaM077 [116]
I think the answer is b
8 0
2 years ago
2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.
hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0
3 years ago
Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the
Nata [24]

<u>Explanation</u>:

For series

\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)

\(I=I_{1}=I_{2}=I_{n}\) (current is the same)

V=I R(\text {voltage is directly proportional to } R)

R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }

For parallel

\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })

I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})

\(I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)\)

\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}

3 0
3 years ago
Need help solving math problem using integration
notka56 [123]
Ummm did you try to add or subtract and multiply or divide that can get your answer
8 0
2 years ago
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0
3 years ago
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