Why Elastic Modulus is important and on what it depends?
1 answer:
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Answer:
1) 1.71 rad/s
2) -6.22 rad/s²
Explanation:
Choose point C to be the origin.
Using geometry, we can show that the coordinates of point A are:
(a cos 30°, a sin 30° − b)
Therefore, the coordinates of point D at time t are:
(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))
The angle formed by CB with the x-axis is therefore:
tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))
1) Taking the derivative with respect to time, we can find the angular velocity:
sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²
We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.
sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²
sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²
dθ/dt = (200) (6) (1/2) / 350
dθ/dt = 600 / 350
dθ/dt = 1.71 rad/s
2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴
At t = 0:
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)
sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)
d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)
Plugging in values:
d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)
d²θ/dt² + 2 tan 30° (1.71) = -4.24
d²θ/dt² = -6.22 rad/s²
Answer:
(a) the percent thermal efficiency is 27.94%
(b) the temperature of the cooling water exiting the condenser is 31.118°C
Explanation:
Answer:
This is the C++ code for the above problem:
#include<bits/stdc++.h>
using namespace std;
int computeFunLevel(int start, int duration, int prefs[], int ngames, int plan[]) {
if(start + duration > 365) { //this is to check wether duration is more than total no. of vaccation days
return -1;
}
int funLevel = 0;
for(int i=start; i<start+duration; i++) { //this loop runs from starting point till
//start + duration to sum all the funlevel in plan.
funLevel = funLevel + prefs[plan[i]];
}
return funLevel;
}
int findBestVacation(int duration, int prefs[], int ngames, int plan[]) {
int max = 0, index = 0, sum = 0 ;
for(int i=1; i<11; i++){ //this loop is to run through whole plan arry
sum = 0; //sum is initialized with zero for every call in plan ,
//in this case loop should run to 366,but for example it is 11
//as my size of plan array is 11
for(int j=0; j<duration; j++) { // this loop is for that index to index+duration to calc
//fun from that index
sum = sum + prefs[plan[i]];
}
if(sum>max) { //this is to check max funlevel and update the index from which max fun can be achieved
max = sum;
index = i;
}
}
return index;
}
int main() {
int ngames = 5;
int prefs[] = { 1,2,0,5,2 };
int plan[] = { 0,2,0,3,3,4,0,1,2,3,3 };
int start = 1;
int duration = 4;
cout << computeFunLevel(start, duration, prefs, ngames, plan) << endl;
cout << computeFunLevel(start, 555, prefs, ngames, plan) << endl;
cout << findBestVacation(4, prefs, ngames, plan) << endl;
}
The screen of the program is given below.
