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lorasvet [3.4K]

3 years ago

10

HI! If you love the art that is good. My teacher Mrs. Armstrong is the best paintings ever year. Come to Mountain View Elementar

y School. Come we have so good of teachers so come. Thank you for reading

2 answers:

Ede4ka [16]3 years ago

5 0

That is a very cool statement I love art very much

QveST [7]3 years ago

4 0

I am too old to go to elementary school :’)

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A 3-m-high, 11-m-wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B. Determine the hyd

serg [7]

Answer:

its a

Explanation:

its a

5 0

3 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

Multiply. Write the answer in simplest form. 1 3/10×1/8

kicyunya [14]

9514 1404 393

Answer:

13/80

Explanation:

The product is ...

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4 0

2 years ago

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3 years ago

Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o

lukranit [14]

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

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The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

7 0

3 years ago