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Natali5045456 [20]
3 years ago
11

Chandler is working on a school editing project. Because he copied 10 GB of raw footage to his computer, its memory is running l

ow. What action will help Chandler create data space in his computer without losing the raw footage?
A.
capture the raw footage on the computer
B.
screen the raw footage on the computer
C.
add effects to the raw footage while editing
D.
log the raw footage and remove the extra footage
Engineering
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

D I think I would of thought about compressing it first tho

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Why or why not the following materials will make good candidates for the construction of
zvonat [6]

Answer:

Answer explained below

Explanation:

3.] a] A turbine blade is the individual component which makes up the turbine section of a gas turbine. The blades are responsible for extracting energy from the high temperature, high pressure gas produced by the combustor.

The turbine blades are often the limiting component of gas turbines. To survive in this difficult environment, turbine blades often use exotic materials like superalloys and many different methods of cooling, such as internal air channels, boundary layer cooling, and thermal barrier coatings. The blade fatigue failure is one of the major source of outages in any steam turbines and gas turbines which is due to high dynamic stresses caused by blade vibration and resonance within the operating range of machinery.

To protect blades from these high dynamic stresses, friction dampers are used.

b] Thermal barrier coatings (TBC) are highly advanced materials systems usually applied to metallic surfaces, such as on gas turbine or aero-engine parts, operating at elevated temperatures, as a form ofexhaust heat management.

These 100μm to 2mm coatings serve to insulate components from large and prolonged heat loads by utilizing thermally insulating materials which can sustain an appreciable temperature difference between the load-bearing alloys and the coating surface.

In doing so, these coatings can allow for higher operating temperatures while limiting the thermal exposure of structural components, extending part life by reducing oxidation and thermal fatigue.

In conjunction with active film cooling, TBCs permit working fluid temperatures higher than the melting point of the metal airfoil in some turbine applications.

Due to increasing demand for higher engine operation (efficiency increases at higher temperatures), better durability/lifetime, and thinner coatings to reduce parasitic weight for rotating/moving components, there is great motivation to develop new and advanced TBCs.

3 0
4 years ago
2. Write a Java program that generates a new string by concatenating the reversed substrings of even indexes and odd indexes sep
Nana76 [90]

Answer:

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        String testString = "abscacd";
  4.        String evenStr = "";
  5.        String oddStr = "";
  6.        for(int i=testString.length() - 1; i >= 0; i--){
  7.            if(i % 2 == 0){
  8.                evenStr += testString.charAt(i);
  9.            }
  10.            else{
  11.                oddStr += testString.charAt(i);
  12.            }
  13.        }
  14.        System.out.println(evenStr + oddStr);
  15.    }
  16. }

Explanation:

Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).

Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.

Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to  testString.length() - 1  (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).

At last, we print the concatenated evenStr and oddStr (Line 18).  

4 0
3 years ago
Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t
Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

end

%This brings an end to the program

4 0
3 years ago
The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.
ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

So

we can say that acceleration= 0.6 x 9.81

acceleration,a=5.88\ \frac{m}{s^2}

We know that

a=\omega ^2A

So now by putting the values

a=\omega ^2A

5.88=\omega ^2 \0.15\times 10^{-3}

\omega =198.09\ \frac{rad}{s}

We know that

  ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

                                               

       

4 0
3 years ago
The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc
balandron [24]

Assuming  the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°)  + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

7 0
2 years ago
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