Answer:
375 KPa
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 125 KPa
Initial temperature (T₁) = 300 K
Final temperature (T₂) = 900 K
Final pressure (P₂) =?
The new (i.e final) pressure of the gas can be obtained as follow:
P₁/T₁ = P₂/T₂
125 / 300 = P₂ / 900
Cross multiply
300 × P₂ = 125 × 900
300 × P₂ = 112500
Divide both side by 300
P₂ = 112500 / 300
P₂ = 375 KPa
Thus, the new pressure of the gas is 375 KPa
Answer:
The average thickness of the blubber is<u> 0.077 m</u>
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Answer:
The following program is in C++.
#include <bits/stdc++.h>
using namespace std;
void lastChars(string s)
{
int l=s.length();
if(l!=0)
{
cout<<"The last character of the string is: "<<s[l-1];
}
}
int main() {
string s;//declaring a string..
getline(cin,s);//taking input of the string..
lastChars(s);//calling the function..
return 0;
}
Input:-
Alex is going home
Output:-
The last character of the string is: e
Explanation:
In the function lastChars() there is one argument that is a string.I have declared a integer variable l that stores the length of the string.If the length of the string is not 0.Then printing the last character of the string.In the main function I have called the function lastChars() with the string s that is prompted from the user.
Answer:
the rate of heat loss by convection across the air space = 82.53 W
Explanation:
The film temperature
to kelvin = (5 + 273)K = 278 K
From the " thermophysical properties of gases at atmospheric pressure" table; At 
= 278 K ; by interpolation; we have the following
→ v 13.93 (10⁻⁶) m²/s
→ k = 0.0245 W/m.K
→ ∝ = 19.6(10⁻⁶)m²/s
→ Pr = 0.713
The Rayleigh number for vertical cavity
= 
= 
For the rectangular cavity enclosure , the Nusselt number empirical correlation:
h = 1.99 W/m².K
Finally; the rate of heat loss by convection across the air space;
q = hA(T₁ - T₂)
q = 1.99(1.4*0.96)(20-(-10))
q = 82.53 W
