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Hunter-Best [27]
3 years ago
11

Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

numbers on one line, each number followed by one space. (2 pts)
Ex:Enter weight 1:236.0Enter weight 2:89.5Enter weight 3:142.0Enter weight 4:166.3Enter weight 5:93.0You entered: 236.000000 89.500000 142.000000 166.300000 93.000000
Engineering
1 answer:
atroni [7]3 years ago
6 0

Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1:  ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2:  ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3:  ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4:  ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5:  ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

format short

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Marizza181 [45]

Answer:

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Explanation:

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3 years ago
To become familiar with the general equations of plane strain used for determining in-plane principal strain, maximum in-plane s
lukranit [14]

Answer:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

Explanation:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

The explanation is shown in the attachment. I hope i have been able to help.

3 0
3 years ago
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm
aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=14.5MPa

T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

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3 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

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I think because if you’ve already turned it in they might as well grade asap instead of waiting
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