Don't break or crush mercury-containing lamps because mercury powder may be released.

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

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter

Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

( $p_t$ )₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ $[$ α₂/( $p_t$ )₂ $]^{1/2$ ------- let this be equation 2

where ( $p_t$ )₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ from equation for (σf)₂ in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ = 2(6( σ₀ )₁) $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

divide both sides by 2(σ₀)₁

$[$ α₁/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]$ = 36 $[$ 0.84/( $p_t$ )₂ $]$

1 / ( $p_t$ )₁ = 30.24 / ( $p_t$ )₂

( $p_t$ )₂ = 30.24( $p_t$ )₁

( $p_t$ )₂/( $p_t$ )₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0

3 years ago

Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t

Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

%This brings an end to the program

4 0

3 years ago

What are the available motor sizes for 2023 ariya ac synchronous drive motor systems in kw?.

Anika [276]

The available motor sizes for 2023 Ariya AC synchronous drive motor systems are:

40 kW.

62 kW.

160 kW.

<h3>What is a synchronous motor?</h3>

A synchronous motor refers to an alternating current (AC) electric motor in which the rotational speed of the shaft is directly proportional (equal) to the frequency of the supply current, especially at a steady state.

In Engineering, the available motor sizes for 2023 Nissan Ariya AC synchronous drive motor systems include the following:

40 kW.

62 kW.

160 kW.

Read more on synchronous motor here:

brainly.com/question/12975042

#SPJ1

5 0

2 years ago

An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The

Mila [183]

Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

6 0

3 years ago