Answer:
Isobaric process
Explanation:
The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.
At constant pressure, the work done is given by :
Where
W is the work done by the system
p is the constant pressure
is the change in volume
So, the correct option is (c) " isobaric process ".
Answer:
A) 0.0614 inches
b) The standard steel paper clip should float on water
Explanation:
The maximum diameter that the rod can have before it will sink
we can calculate this using this formula :
D = ----- 1
∝ = value of surface tension of water at 60⁰f = 5.03×10^−3 lb/ft
y = 490 Ib/ft^3
input the given values into equation 1 above
D =
= 5.11 * 10^-3 ft convert to inches
= 5.11 *10^-3 ( 12 in/ 1 ft ) = 0.0614 inches
B) The diameter of a standard paper Cliphas = 0.036 inches
and the diameter of the rod = 0.0614. Hence the standard steel paper clip should float on water
commutative property of addition
identity property of multiplication
associative property of addition
commutative property of multiplication
The property of realnumbers is shown below is associative property of addition. The correct option is C.
According to the associativeproperty of addition, you can arrange the addends in several ways without changing the result.
According to the commutative property of addition, you can rearrange the addends without altering the result.
When more than two numbers are added together or multiplied, the outcome is always the same, regardless of how the numbers are arranged.
This is known as the associativeproperty. As an illustration, 2 (7 6) = (2 7) 6. 2 + (7 + 6) = (2 + 7) + 6.
Thus, the correct option is C.
For more details regarding associative property, visit:
#SPJ1
Answer:
C
Explanation:
Buffering is an important op amp application because it solves impedance-matching problems that can't easily be solved with purely resistive circuits. This allows for the transfer of maximum power between two circuits without any loss of signal strength.
Buffering with operational amplifiers (op amps) is a crucial application in electronics because it helps to overcome impedance-matching problems that cannot be easily resolved with only resistive circuits. Impedance-matching issues can cause signal distortion, and buffering solves this problem by creating a high-input impedance and low-output impedance circuit that separates the input and output signals, preventing the signal from being affected by the load impedance.
By using op amp buffering, the signal can be efficiently transmitted and received without signal loss or distortion, making it a useful technique in many applications, such as audio amplification and signal conditioning.
"
Complete question
Buffering is an important op amp application because it solves _____ that can't easily be solved with purely resistive circuits.
Group of answer choices
delay issues
tolerance issues
temperature issues
loading effects
impedance-matching problems
"
You can learn more about Buffering at
#SPJ11
Answer:
See attached pictures.
Explanation:
See attached pictures for detailed explanation.
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;
}
In this exercise, using the knowledge of computational language in C++, we have that this code will be written as:
The code is in the attached image.
We can write the C++ as:
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int divSum(int num)
{
int result = 0;
for (int i=2; i<=(num/2); i++)
{
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;
}
See more about C++ at brainly.com/question/19705654