Answer:
The correct option is;
A. be in compliance with school attendance requirements
Explanation:
The requirements to acquire a special restricted driver's license for driver's license for under under 18 that have had a beginner's for up to 180 days include a minimum of 40 hours driving practice 10 of which should be in the dark. The application for the conditional drivers license is to be signed by the parent or guardian and the application is to be accompanied with proof of acceptable school attendance
At 17, after holding the special restricted drivers licence for a year without issues you can obtain the full drivers license.
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.
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
end
%This brings an end to the program
Answer:
b. 10A
Explanation:
Using the formula, E= k × r×I
200= 0.5 ×2000×0.02×I
200=20×I
Dividing with 20
I = 200/20= 10A
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
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<em>attached below is the detailed solution </em>
