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3 years ago

7

Write a MATLAB statement that results in the input request shown in bold. The >> shows where your statement is typed, an

d the | shows where the cursor waits for input. The display must be correctly positioned. Each has a space before the cursor (shown as |). The input variable name and the variable type are shown at the right.

1 answer:

statuscvo [17]3 years ago

3 0

Answer:

For the given input request, the MATLAB statement is as follows

Bolt=input(‘Enter the length of the bolt in inches:’);

Consequently, the statement is written

The next MATLAB statement is

Bolt=input(‘Enter the company’s name: ‘s’);

The statement is therefore, written

Finally

Line_color=menu(‘Choose a color’, ‘Red’, ‘Green’, ‘Blue’);

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Calculate the energy in kJ added to 1 ft3 of water by heating it from 70° to 200°F.

Serga [27]

Answer:

The energy in kJ is 8558.16 kJ.

Explanation:

Data presented in the problem:

Water is heated from 70 (T1) to 200 °F (T2).

Volume (V) of the water is 1 ft3.

It is required for the specific heat of water(HW), which is 1 BTU/lb°F.

First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.

M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.

.After that, we can calculate the heat required (Q).

Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)

Q = 62.4 * 130 BTU = 8112 BTU.

Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ

Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.

Finally, the energy required is 8558.16 kJ.

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3 years ago

Ok this really isn’t a question but I need help, I’m wondering if a Samsung galaxy 9 is a good phone

Savatey [412]

Answer:

yes it is

Explanation:

it's an extremely good phone in my opinion

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3 years ago

Read 2 more answers

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

A cylindrical metal specimen having an original diameter of 10.55 mm and gauge length of 54.5 mm is pulled in tension until frac

kumpel [21]

Answer:

(a) 53.94%

(b) 26.61%

Explanation:

Change in area will be given by

$\triangle A=\frac {\pi(R_o^{2}-r_n^{2})}{\pi R_o^{2}}$ where $\triangle A$ represent change in area R is radius and subscripts O and n represent original and new respectively.

Substituting 10.55/2 for original radius and 7.16/2 for new radius then

$\triangle A=\frac {\pi(5.275^{2}-3.58^{2})}{\pi 5.275^{2}}\times 100\approx 53.94$

(b)

Similarly, percentage elongation will be found by dividing the change in length by the the original length. In this case, rhe original length was 54.5mm and goes to 69 mm so the change in length is given by subtracting the final length from the original length

Percentage elongation is $\frac {69-54.5}{54.5}\times 100\approx 26.61$

6 0

4 years ago

Indicate whether the following statements are true or false for an isothermal process: (A) Q=T(∆S). (B) ∆U=0.(C) The entropy cha

Lena [83]

Answer:

A=False

B=False

C=False

D=False

E=False

F=False

Explanation:

A. In an isothermal process, only the reversibly heat transfer is 0, $Q_{rev}=T (\Delta S)$

B. Consider the phase change of boiling water. Here, the temperature remains constant but the internal energy of the system increases.

C. This is not true even in reversible process, as can be inferred from the equation in part A.

D. This is only true in reversible processes, but not in all isothermal processes.

E. Consider the phase change of freezing water. Here, the surroundings are increasing their entropy, as they are taking in heat from the system.

F. This is not true if $(\Delta U)\neq 0$ , like in answer B. One case where this is true is in the reversible isothermal expansion (or compression) of an ideal gas.

3 0

3 years ago