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lesantik [10]
3 years ago
6

In order to lift a lighter object on the other side, a boy placed 155 N of

Engineering
1 answer:
const2013 [10]3 years ago
4 0

Answer:

0.58 m

Explanation:

In order to find the vertical distance the load lifts, we first need to find the magnitude of the resistance force. So, Rd = Fd' where R = resistance force = ?, d = resistance distance = 3.5 m, F = effort force = 155 N and d' = effort distance = 1.5 m.

So R = Fd'/d = 155 N × 1.5 m/3.5 m = 232.5 Nm/3.5 m = 66.43 N  

Now, the vertical work done by the effort = vertical work done by load

Fy = Ry' where y = vertical distance moved by effort = 0.25 m and y' = vertical distance moved by resistance force

So, Fy = Ry'

y' = Fy/R

= 155 N 0.25 m/66.43 N

= 38.75 Nm/66.43 N

= 0.58 m

So, the vertical distance the lighter object moves is y' = 0.58 m

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Air enters the compressor of a regenerative gas-turbine engine at 300 K and 100 kPa, where it is compressed to 800 kPa and 580 K
Bad White [126]

Answer:

a) The amount of heat transfer in the regenerator, q = 114.12 kJ/kg

b) Thermal efficiency = 35.9%

Explanation:

The calculations are neatly handwritten and attached as files to this solution for easiness of expression and clarity. The cycle is also drawn on the T - S diagram and included in the attached files. Check the files below for the complete calculation.

I hope this helps!

3 0
3 years ago
Design an algorithm for computing √n
a_sh-v [17]
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8 0
2 years ago
A rotating shaft is subjected to a steady torsional stress of 13 ksi and an alternating bending stress of 22 ksi.
mixas84 [53]

Answer:

A) б1 = 28 ksi and  б2 = -6.02 ksi

B) 1.25

Explanation:

Given data :

Torsional stress = 13 ksi

Alternating bending stress = 22ksi

A) determine yielding factor of safety  according to the distortion energy theory

б1,2 = \frac{22}{2} ± √(22/2)² + 13²

       = 11  ± 17

therefore б1 = 28 ksi  hence б2 = -6.02 ksi

B) determine the fatigue factor of safety  

with properties ;  Se = 35ksi, Sy = 60 ksi, Sut = 85 ksi

( б1 - б2 )²  + ( б2 - б3 )² + ( б3 - б1 )²  ≤  2 ( Sy / FOS ) ²

( 28 + 6.02 ) ² + ( 6.02 - 0 )² + ( 0 - 28 )² ≤  2 ( 60 / FOS ) ²

solving for FOS = 1.9

Next we can determine FOS with the use of Goodman criterion

бm / Sut  + бa / Se  =  1 / FOS

= 0 / 85 + 28/35 = 1 / FOS

making FOS the subject of the equation ; hence  FOS = 1.25

3 0
3 years ago
This question is in C programming.Write a loop that sets newScores to oldScores shifted once left, with element 0 copied to the
andriy [413]

Answer:

Replace

/* Your solution goes here */

with the following;

for (i = 0; i < SCORES_SIZE - 1; i++) {

newScores[i] = oldScores[i + 1];

}

newScores[SCORES_SIZE - 1] = oldScores[0];

The full program becomes

#include

int main(void) {

const int SCORES_SIZE = 4;

int lowerScores[SCORES_SIZE];

int i;

for (i = 0; i < SCORES_SIZE; ++i) {

scanf("%d", &(lowerScores[i]));

}

for (i = 0; i < SCORES_SIZE - 1; i++) {

newScores[i] = oldScores[i + 1];

}

newScores[SCORES_SIZE - 1] = oldScores[0];

for (i = 0; i < SCORES_SIZE; ++i) {

printf("%d ", lowerScores[i]);

}

printf("\n");return 0;

}

What the included code segment does it that;

for (i = 0; i < SCORES_SIZE - 1; i++) {

newScores[i] = oldScores[i + 1];

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The above assigns the values of array oldScores to newScores starting with the second element of array oldScores till the last.

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The above line then assigns the first element of oldScores to the first element of newScores array

7 0
4 years ago
A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium
Juli2301 [7.4K]

Answer:

N_A=1.5*10^-8 kmol/s.m^2

Explanation:

<u>KNOWN: </u>

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.  

<u>FIND:</u>

Molar diffusion flux.  

<u>ASSUMPTIONS:</u>

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.  

<u>ANALYSIS:</u> The molar flux may be obtained from

 N_A=D_AB/L(C_A,1-C_A,2)

       =10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3

N_A=1.5*10^-8 kmol/s.m^2

<u>COMMENTS:</u> The mass flux is:

n_A,x=M_a*N_A,x

n_A,x=6*10^-8 kg/s m^2

5 0
3 years ago
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