Answer:
Explanation:
,K=1.35
Clearance is 8%.
Heat added=15 KJ
We know that compression ratio
r=13.5
We know that efficiency of otto cycle
W is the work out put and Q is the heat addition.
W=8.4 KJ
We know that Work =Mean effective pressure x swept volume.
Here swept volume
Noe by putting the values
Work =Mean effective pressure x swept volume.
Answer:
apparent position = 26.6 cm
and size is same of coin because we observe it directly above of the coin
Explanation:
given data
diameter = 2 cm
depth = 20 cm
refractive index of water = 1.33
to find out
apparent position and size
solution
we know refractive index of air is 1
so apparent position of coin formula is
apparent position = ..........................1
put here value in equation 1
apparent position =
apparent position = 26.6 cm
and size is same of coin because we observe it directly above of the coin
Answer:
No
Explanation:
He is actually still alive
Answer:
the heat transfer from the pipe will decrease when the insulation is taken off for r₂<
where;
r₂ = outer radius
= critical radius
Explanation:
Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .
The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< ) 0, and reaches a maximum when r₂ = , and starts to decrease for r₂< . Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< .
Answer:
Explanation:
/* General Code for achieving perfect error for any level and any
digit of codewords*/
#include<iostream>
using namespace std;
int main(){
int level;
//Input memory levels.
cout <<" Enter number of Flash Memory levels : ";
cin>>level;
//construct array store levels.
int lvl[level];
for(int i = 0; i < level; i++){
//assign levels name
lvl[i] = i;
}
int digit;
cout << "Enter number of digit codewords :";
//Taking input of codeword digits
cin >> digit;
for(int i = 0; i < level;){
//output Codeword
cout << lvl[i];
//Skipping for n number digit codeword in level
i+=(digit+1);
}
}
/*
a) i from 0 to (i = i+(2+1))<levels
b) for length (1) -> 3 codewords
codewords -> {0,2,4}
for length (2) -> 2 codewords
codewords -> {0,3}
*/