The moons light-colored highlights are

A rigid bar ABCD is pinned at A and supported by two steel rods connected at B and C, as shown. There is no strain in the vertic

mylen [45]

Answer:

See attached picture.

Explanation:

4 0

3 years ago

For this homework you are writing a main function, with 3 subfunctions. 2. The "main" function is to be named arm_prop a. Inputs

ra1l [238]

Answer:

MATLAB CODE:

clc;

clear all;

x=[1 2 3 4]; %Input vector of variable size

[gm,ra,hm]=arm_prop(x) %Main function calling , output values are hm, ra, hm and imput is x

%% Main function

function [gm,ra,hm]=arm_prop(x)

[gm]=geo_mean(x); %Calling Sub function 1

[ra]=rms_avg(x); %Calling Sub function 2

[hm]=har_mean(x); %Calling Sub function 3

%Sub function 2 =Rms average

function [gm]=geo_mean(x)

n=numel(x); %calculating number of elemnts in vector x

temp=1; %Temporary value for multiplying each and every value of x

for i=1:1:n % For loop run for number of elements in x

temp=temp*x(i); %Multliplyng each and every element of x

end

gm=temp^(1/n); % Calculating geometric mean

end

%Sub function 2 =Rms average

function [ra]=rms_avg(x)

temp=0; %Temporary value for adding square of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+x(i)^2; %Adding square of each and every element of x

end

ra=(temp/N)^(1/2); % Calculating Rms average

end

%Sub function 3 =Harmonic mean

function [hm]=har_mean(x)

temp=0; %Temporary value for adding reciprocal of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+(1/x(i)); %Adding sreciprocal of each and every element of x

end

hm=N/temp; % Halculating Harmonic mean

end

Results:

gm =

2.2134

ra =

2.7386

hm =

1.9200

Here in one main function. 3 sub functions are written. Every sub function returns one output. As all 3 sub functions are inside main function, main function returns 3 output values gm, ra, hm.

Explanation:

3 0

4 years ago

Two forces, one of which double the other has resultant of 280N. if the direction of the large force is reversed and the remaini

White raven [17]

Answer:

point is 16 N.If their resultant is normal to the smaller force and has a magnitude of 8 N. Then two forces are.

4 0

4 years ago

The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us

devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.

This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function $e^{j(wt+\alpha)}$ ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0

3 years ago

You need to wear respiratory PPE whenworking around materials that may contain asbestos

choli [55]

Answer:

to prevent asbestos fibers poisoning

Explanation:

There is a potential for asbestos fibers to be released when removing asbestos-containing material,using personal protective equipment will significantly reduce your exposure to asbestos fibers.

7 0

2 years ago