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3 years ago

6

Create a file "parts_inv.dat" that stores on each line a part number, cost, and quantity in inventory, in the following format:1

23 5.99 52Use fscanfto read this informationand print the total dollar amount of inventory (the sum of the cost multiplied by the quantity for each part).

1 answer:

andriy [413]3 years ago

7 0

Answer:

%Open the file.

fID = fopen('parts_inv.dat');

%Read from the file.

data = fscanf(fID,'%d\t%f\t%d',[3,inf]);

%Close

fclose(fID);

%Restore the data.

data = data';

%Get the size

[rs, cs] = size(data);

%Set value.

invCost = 0;

%Loop

for rw = 1 : rs

%Find cost

invCost = invCost + (data(rw, 2) * data(rw, 3));

%Loop end

end

%Display the cost.

fprintf('Total cost: %4.2f\n\n', invCost);

Explanation:

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If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.4 m/s how much work does sam do to stop it?

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A satellite s is moving in an elliptical orbit around the earth . the mass of satellite is very small compared to mass of earth.

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3 years ago

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

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3 years ago

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

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3) for heated metal T = 1500 K

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$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

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3 years ago