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jarptica [38.1K]
2 years ago
8

What is the law of physics 10 points if you tell me the answer and your name

Engineering
1 answer:
levacccp [35]2 years ago
7 0
<h2>꧁༒Answer༒꧂</h2>

<h2>F=m•a</h2>

f = force

m = mass \: of \: an \: object

a = acceleration

Newton's second law is one of the most important in all of physics. For a body whose mass m is constant, it can be written in the form F = ma, where F (force) and a (acceleration) are both vector quantities. If a body has a net force acting on it, it is accelerated in accordance with the equation

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¿Qué áreas del conocimiento me pueden<br> aportar a la ejecución del proyecto?
allsm [11]

Answer:

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3 years ago
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
3 years ago
Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a numb
tekilochka [14]

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;

}

6 0
3 years ago
Which examples demonstrate tasks commonly performed in Maintenance/Operations jobs? Check all that apply.
Verdich [7]

1.Ross fixes a dishwasher for a homeowner.

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4 0
3 years ago
Read 2 more answers
Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces
RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

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Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0
3 years ago
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