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nexus9112 [7]
2 years ago
11

The sum of forces on node 2 (upper-left) is ______.

Engineering
1 answer:
laiz [17]2 years ago
6 0

In the case above, the sum of forces on node 2 (upper-left) is said to be zero.

<h3>What is net force?</h3>

If any object is said to be falling at constant speed. The net force is often known to be  zero.

in the case above, you see that 2 and 3 node is falling down therefore, node 1 and 4 will be zero.

So, In the case above, the sum of forces on node 2 (upper-left) is said to be zero.

Learn more about net force from

brainly.com/question/11556949

#SPJ1

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Explanation:

sorry its too late tho

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Orbit is to _____ as altitude is to _____.
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Explanation:

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LAB 3.3 – Working with String Input and Type CastingStep 1: RemovefindErrors.cppfrom the project and add thepercentage.cppprogra
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Answer:

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

#include<iostream>

using namespace std;

int main ()

{

// Variable declaration

string name;

int numQuestions;

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double percentage;

//Prompt to enter student's first and last name

cout<<"Enter student's first and last name";

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Explanation:

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4 years ago
Using the data in the photo write the complex waveform expression​
UNO [17]

Answer:

1st Harmonic:

v(t) = 50\cos(2000\pi t)

3rd Harmonic:

v(t) = 9\cos(6000\pi t)

5th Harmonic:

v(t) = 6\cos(10000\pi t)

7th Harmonic:

v(t) = 2\cos(14000\pi t)

Explanation:

The general form to represent a complex sinusoidal waveform is given by

v(t) = A\cos(2\pi f t + \phi)

Where A is the amplitude in volts of the sinusoidal waveform

Where f is the frequency in cycles per second (Hz) of the sinusoidal waveform

Where \phi is the phase angle in radians of the sinusoidal waveform.

1st Harmonic:

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3rd Harmonic:

We have A = 9, f = 3000 and φ = 0

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We have A = 6, f = 5000 and φ = 0

v(t) = 6\cos(2\pi 5000 t + 0) \\\\v(t) = 6\cos(10000\pi t)

7th Harmonic:

We have A = 2, f = 7000 and φ = 0

v(t) = 2\cos(2\pi 7000 t + 0) \\\\v(t) = 2\cos(14000\pi t)

Note: The even-numbered harmonics have 0 amplitude that is why they are not shown here.

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