Answer:
Increases
Explanation:
By inhibiting the motion of dislocations by impurities in a solid solutions, is a strengthening mechanism. In solid solutions it is atomic level strengthening resulting from resistance to dislocation motion. Hence, the strength of the alloys can differ with respect to the precipitate's property. Example, the precipitate is stronger (ability to an obstacle to the dislocation motion) than the matrix and it shows an improvement of strength.
Answer: c) they have low genetic variability among them.
When a plant is grown for several generations of offspring of a plant, then there are some common things which are to be noted which are found similar in the offspring and in the parent of the offspring. The flowers and fruits and the time or season they come in are absolutely the same.
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer:
// Program is written in C++
// Comments are used to explain some lines
// Only the required function is written. The main method is excluded.
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int divSum(int num)
{
// The next line declares the final result of summation of divisors. The variable declared is also
//initialised to 0
int result = 0;
// find all numbers which divide 'num'
for (int i=2; i<=(num/2); i++)
{
// if 'i' is divisor of 'num'
if (num%i==0)
{
if (i==(num/i))
result += i; //add divisor to result
else
result += (i + num/i); //add divisor to result
}
}
cout<<result+1;
}
