Entropy change of hot and cold reservoir are -0.08333 kJ/k and 0.16666 kJ/k respectively.
<u>Explanation</u>:
Given:
Temperature at hot reservoir,
Temperature at cold reservoir,
Amount oh heat transferred,
Entropy change at hot reservoir,
Entropy change at cold reservoir,
Total entropy change,
is not less than zero.Hence,it fulfills increase of entropy principle.
Answer:
ohm
Explanation:
Given data:
capacitor = 0.5 micro farad
resistor = 500 ohm
Frequency = 2000 Hz
Impedance of any circuit is calculated by using following equation
=500 - j 159.155
ohm
Answer:
c. multiple trips to many specialists.
Explanation:
These are a organizations that provide health insurance time fee.
Answer:
Explanation:
given data
Load P = 35 kN
Width of bar W = 50.8 mm
Breadth of bar B = 25 mm
Ratio of crack length to width α = a/W = 0.2
solution
we get here KI for a rectangular bar that is express as
................................1
here Y is the geometrical function
so
Y =
Y =
Y =
Y = 0.9878
so put here value in equation 1
= 5210.45 × 10³
= 5.21 MPa
Answer:
See explaination and attachment for theprogram code and output.
Explanation:
matlab program:
x1(1)=0;
y1(1)=0;
x2(1)=0;
y2(1)=0;
x3(1)=0;
y3(1)=0;
for i=1:10
x1(i+1)=0.5*x1(i);
y1(i+1)=0.5*y1(i);
x2(i+1)=0.5*x2(i)+0.25;
y2(i+1)=0.5*y2(i)+(sqrt(3)/4);
x3(i+1)=0.5*x3(i)+0.5;
y3(i+1)=0.5*y3(i);
end
figure
hold on
plot(x1,y1,'*')
plot(x2,y2,'*')
plot(x3,y3,'*')
hold off
Please kindly check attachment for output of 10, 100 and 1000 respectively.