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JulijaS [17]
3 years ago
13

The opening of the throttle plate can be delayed as long as

Computers and Technology
1 answer:
blsea [12.9K]3 years ago
6 0
<h2>I'm trying to adjust the sawtooth throttle on my H. I can open the throttle to 8 notches from the top and it is at the 1815 rated rpm. When I go above that it will overspeed, up to 2500rpm! At the last notch on the quadrant. I have tried the adjustments in the governor, the only thing I have been able to do is set the high rpm screw so it stops at the 8th notch from the top. I should be able to have full use of the quadrant and not go over the 1815 rpm, right? Does anyone have a pic of the throttle linkage under the hood and how to adjust it? I've also tried adjusting the rod in the governor between it and the carb, it is tightened all the way up now. Any ideas what I am missing?</h2>
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The height of a small rocket y can be calculated as a function of time after blastoff with the following piecewise function: y 5
SOVA2 [1]

Answer:

High level Language

understand

Explanation:

rocket is 0...4433456u888

5 0
2 years ago
Pa sagot po TLE-10<br>need den po Ng solution thank you​
horsena [70]

Answer:

HAHHAHAHA HAAHA g abbabanjaja abunjing <em><u>abunjin</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em>

3 0
2 years ago
Write a program that simulates flipping a coin repeatedly and continues until three consecutive heads. are tossed. At that point
quester [9]

Answer and Explanation:

#include <iostream>

#include <string>

#include <time.h>

#include <vector>

using namespace std;

//Takes user info

int user_info(string *name,char *gender){

   char G;

   cout<<"What is your name?: ";

   getline(cin, *name);

   cout<<"What is your gender(m/f):";

   cin>>G;

   cout<<endl;

   *gender = tolower(G);

   return 0;

}

//Toss coin(part 2)

bool coin_toss(double seed){

   double r;

   r = (double)seed/(double)RAND_MAX;

   //cout<<r;

   if (r<0.5) return true;

   return false;

}

//check results

float toss_result(vector<char> v,int *h){

   int num_heads=0;

   int num_toss = v.size();

   *h = 0;

   for(int i=0;i<num_toss;i++){

       if(v[i]=='h'){num_heads++;(*h)++;}

   }

   return (float)num_heads/(float)num_toss;

}

void show_result(int total_num,int head_num,float ratio){

   cout<<"it took you "<<total_num<<" tosses to get 3 heads in a row"<<endl;

   cout<<"on average you flipped heads "<<(ratio*100)<<"% of the time"<<endl;

}

int main()

{

   string name;

   char gender;

   string K; //Mr or Mrs

 

   cout<<"Welcome to coin toss! Get 3 heads in row to win!"<<endl;

 

   //part 1

   user_info(&name,&gender);

 

   if(gender == 'f') K = "Mrs.";

   else K = "Mr.";

 

   cout<<"Try to get 3 heads in a row. Good luck "<<K<<name<<"!"<<endl;

   char dummy = getchar();

 

   //part 2

   int num_toss;

   vector<char> toss_results;//store toss results

   bool result;//result of toss

   srand(time(0));

   while(true){

       //cout<<rand()<<endl;

       result = coin_toss(rand());

       cout<<"Press <enter> to flip"<<endl;

       while (1)

       {

           if ('\n' == getchar())

           break;

       }

     

 

     

       if(result) {toss_results.push_back('h');cout<<"HEAD"<<endl;}

       else {toss_results.push_back('t');cout<<"TAIL"<<endl;}

     

       num_toss = toss_results.size();

     

       if(num_toss>=3){

           if ((toss_results[num_toss-1] == 'h')&&(toss_results[num_toss-2] == 'h')&&(toss_results[num_toss-3] == 'h')){

               break;

           }

       }

   }

 

   //part 3

   float ratio_head;

   int num_of_heads;

   ratio_head = toss_result(toss_results,&num_of_heads);

 

   //part 4

 

   show_result(toss_results.size(),num_of_heads,ratio_head);

 

   return 0;

}

8 0
3 years ago
What is the output of the following code segment?
Aliun [14]

Answer:

o

Explanation:

8 0
2 years ago
A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on average once every 100 sec, e
Sunny_sXe [5.5K]

Explanation:

Here, in the given statement, maximum flow capacity of the ALOHA channel = 18.4% = 0.184

Then, the stations N share 56 kbps

And, the channel rate = 0.184*56 = 10.304 kbps.  

Then, the outputs of the wach stations is 1000-bit frames/100 sec

\therefore, the output of the bits/ sec by each stations = \frac{1000}{100} bits/sec = 10 bits/sec

So, The output of the N stations is 10 bits/sec on the channel that having 10.304kb/sec

Finally, N = \frac{(10.304 \times 10^{3})}{10} = 1030 stations

3 0
3 years ago
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