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Crank
3 years ago
13

Question 2

Engineering
1 answer:
dimaraw [331]3 years ago
6 0

Answer:

[tex]\blue{\rule{200pt}{555559999955pt}[\tex]

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I will give Brainliest, please help. :)
ziro4ka [17]

Answer:

The ratio of water to flour needed to make tortillas, which is 3 : 4

Explanation:

100 w = 75 f

w/f = 75/100 = 3/4

7 0
3 years ago
Read 2 more answers
Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa
ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

%Program shows the value that user entered

fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

   %check if the element is a positive number

   if v(i) > 0

       %double the element

       v(i) = v(i) * 2;

   %else the element is negative number.

   else

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       v(i) = v(i) * 3;

   end

end

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fprintf('The modified vector:\n ')

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4 0
4 years ago
A(n) _____ is an apparatus that changes alternating current (AC) to direct current (DC)
amid [387]

Answer:

rectifier

Explanation:

7 0
2 years ago
What was the purpose of the vasa ship
goldfiish [28.3K]
The main purpose was for power. The vessel has come to symbolize Sweden's Great Power Period, when the nation became a major European power and controlled much of the Baltic.
8 0
3 years ago
If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical
vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 ×10^{6} N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 ×10^{9} N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}    .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm  = 0.3 ×10^{-3} m

so now put value in equation 1 we get

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}

( σc ) = \sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}

( σc ) = 27.396615 ×10^{6} N/m²

so critical stress required for the propagation is 27.396615 ×10^{6} N/m²

6 0
3 years ago
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