Answer:
The code will be:
#include <stdio.h>
#include <stdlib.h>
main () {
double weight, shippingCharge, rate, segments;
int distance;
printf("Enter the weight: \n");
scanf("%lf", &weight);
printf("Enter the distance: \n");
scanf("%i", &distance);
if (weight <= 10) {
printf("Rate is $3.00 \n");
rate = 3;
} else {
printf("Rate is $5.00 \n");
rate = 5;
}
if (distance % 500 == 0) {
segments = distance / 500;
} else {
segments = distance / 500 + 1;
}
shippingCharge = rate * segments;
if (distance >1000) {
shippingCharge = shippingCharge + 10;
}
printf("Your shipping charge is $%lf\n", shippingCharge);
system ("pause");
}
Explanation:
An XOR gate (sometimes referred to by its extended name, Exclusive OR gate) is a digital logic gate with two or more inputs and one output that performs exclusive disjunction. The output of an XOR gate is true only when exactly one of its inputs is true.
Answer:well u can use to make a shelter but that's all I can think of ??
Explanation:
Answer:
0.024 m = 24.07 mm
Explanation:
1) Notation
= tensile stress = 200 Mpa
= plane strain fracture toughness= 55 Mpa
= length of a surface crack (Variable of interest)
2) Definition and Formulas
The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.
By definition we have the following formula for the tensile stress:
(1)
We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for 
Multiplying both sides of equation (1) by 
(2)
Sequaring both sides of equation (2):
(3)
Dividing both sides by
we got:
(4)
Replacing the values into equation (4) we got:
![\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5B%5Cfrac%7B55%20Mpa%5Csqrt%7Bm%7D%7D%7B1.0%28200Mpa%29%7D%5D%5E2%20%3D0.02407m)
3) Final solution
So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432