Answer:
shear plane angle Ф = 26.28°
shear strain 2.20
Explanation:
given data
angle = 16°
chip thickness t1 = 0.32 mm
cut yields chip thickness t2 = 0.72 mm
solution
we get here first chip thickness ratio that is
chip thickness ratio =
................. 1
put here value
chip thickness ratio =
chip thickness ratio r = 0.45
so here shear angle will be Ф
tan Ф =
............2
tan Ф =
tan Ф = 0.4938
Ф = 26.28°
and
now we get shear strain that is
shear strain r = cot Ф + tan (Ф - α ) ................3
shear strain r = cot(26.28) + tan (26.28 - 16 )
shear strain r = 2.20
Answer:
clc
clear
x = input('type value of angle in degrees:\n');
x = x*pi/180; %convverting fron degree to radian
sin_x = x; %as first term of taylor series is x
E = 1; %just giving a value of error greater than desired error
n = 0;
while E > 0.000001
previous = sin_x;
n = n+1;
sin_x = sin_x + ((-1)^n)*(x^(2*n+1))/factorial(2*n+1);
E = abs(sin_x - previous); %calculating error
end
a = sprintf('sin(x) = %1.6f',sin_x);
disp(a)
Explanation:
Answer:
get mutual funds
Explanation:
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