Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
%Program prompts user to input vector
v = input('Enter the input vector: ');
%Program shows the value that user entered
fprintf('The input vector:\n ')
disp(v)
%Loop for checking all array elements
for i = 1 : length(v)
%check if the element is a positive number
if v(i) > 0
%double the element
v(i) = v(i) * 2;
%else the element is negative number.
else
%triple the element
v(i) = v(i) * 3;
end
end
%display the modified vector
fprintf('The modified vector:\n ')
disp(v)
Answer:
The current drawn from the outlet is 0.2 A
The number of turns on the input side is 350 turns
Explanation:
Given;
number of turns of the secondary coil, Ns = 35 turns
the output current, 
= 2 A
power supplied, 
= 24 W
the standard wall outlet in most homes = 120 V = input voltage
For an ideal transformer; output power = input power
the current drawn from the outlet is calculated;
The number of turns on the input side is calculated as;
