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");
}
Answer:
you get electrocuted...........
Answer:
hi-he = 0
pi-pe = positive
ui-ue = negative
ti-te = negative
Explanation:
we know that fir the sub cool liquid water is
dQ = Tds = du + pdv ............1
and Tds = dh - v dP .............2
so now for process of throhling is irreversible when v is constant
then heat transfer is = 0 in irreversible process
so ds > 0
so here by equation 1 we can say
ds > 0
dv = 0 as v is constant
so that Tds = du .................3
and du > 0
ue - ui > 0
and
now by the equation 2 throttling process
here enthalpy is constant
so dh = 0
and Tds = -vdP
so ds > 0
so that -vdP > 0
as here v is constant
so -dP =P1- P2
so P1-P2 > 0
so pressure is decrease here
Answer: The net force in every bolt is 44.9 kip
Explanation:
Given that;
External load applied = 245 kip
number of bolts n = 10
External Load shared by each bolt (P_E) = 245/10 = 24.5 kip
spring constant of the bolt Kb = 0.4 Mlb/in
spring constant of members Kc = 1.6 Mlb/in
combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in
Initial pre load Pi = 40 kip
now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them
External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip
So Total net Force on each bolt Fb = P_Eb + Pi
Fb = 4.9 kip + 40 kip
Fb = 44.9 kip
Therefore the net force in every bolt is 44.9 kip
Answer:
q=2313.04
T=690.86°C
Explanation:
Given that
Thickness t= 20 cm
Thermal conductivity of firebrick= 1.6 W/m.K
Thermal conductivity of structural brick= 0.7 W/m.K
Inner temperature of firebrick=980°C
Outer temperature of structural brick =30°C
We know that thermal resistance
These are connect in series
Heat transfer
So heat flux
q=2313.04
Lets temperature between interface is T
Now by equating heat in both bricks
So T=690.86°C
