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nevsk
10 hours ago
15

Open the"stateData3.c" program and try to understand how the tokenization works. If you open the input file "stateData.txt", you

can clearly seethat a comma separates the state name and its population. The tokenizer portion of the programs separates each line into two tokens and stores them into two different arrays. You need to display the arrayelements such that each line has a state and itspopulation. Also calculate the total population of USA.Then,examine the portion of the code how it writes the array elements into a binary datafile.You need to write similar logic where it writes the array elements into a text file "stateDataOutput2.txt". Please check the syntax and usage of fprintf(); and use that here.
Engineering
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A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr
choli [298]

Answer:

you may be struggling to pinpoint the separation between your inquiry and my perspective

0 0
3 months ago
Write cout statements with stream manipulators that perform the following:
grin007 [323]

Answer:

A)cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B)cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C)cout<<fixed<<5.789E12;

D)cout<<left<<setw(7)<<67;

Explanation:

Stream Manipulators are special functions for use with the insertion (<<) and extraction (>>) operators on C++ stream objects, while the 'cout' statement outputs content to the standard output device in C++ programming.

setw: specifies the minimum width of the output field

setprecision: defines the number of decimal places for floating-point value formatting.

fixed: sets the format flag for floating-point streams.

left: left-aligns the output.

A) This statement shows the number 34.789 in a field that provides eight character spaces with two decimal precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B) Here, the number 7.0 is displayed within six spaces with three decimal precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C) This command prints 5.789e+12 in fixed-point format.  cout<<fixed<<5.789E12;

D) This statement left-aligns the number 67 across a field of six spaces. cout<<left<<setw(7)<<67;

7 0
3 months ago
The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re
mote1985 [299]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, g = 9.81 m/s^2

mass flow rate, \dot{m} = 60.6 kg/s

flow density, \rho = 1000 kg/m^3

efficiency of the pump, \eta = 0.74

output gauge pressure, p_o = 344.75 kPa

input gauge pressure, p_i = 68.95 kPa

cross-sectional area of output pipe, A_o = 0.069 m^2

cross-sectional area of input pipe, A_i = 0.093 m^2

height of discharge, z_o = 1.22 m - 0.61 m = 0.61 m (evaluated at pump’s maximum height of 1.22 m)

input height, z_i = 0 m

hydraulic power of the pump,P =? kW

Initially, the volumetric flow (Q) needs to be determined

Q = \frac{\dot{m}}{\rho}

Q = \frac{60.6 kg/s}{1000 kg/m^3}

Q = 0.0606 m^3/s

Next, compute the velocity (v) for both input and output

v_o = \frac{Q}{A_o}

v_o = \frac{0.0606 m^3/s}{0.069 m^2}

v_o = 0.88 m/s

v_i = \frac{Q}{A_i}

v_i = \frac{0.0606 m^3/s}{0.093 m^2}

v_i = 0.65 m/s

Subsequently, the total head (H) can be calculated

H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}

H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}

H = 28.74m

Finally, the computation of pump power is done as follows

P = \frac{Q \, \rho \, g \, H}{\eta}

P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}

P = 23.09 kW

6 0
3 months ago
The molecular weight of a 10g rubber band
Viktor [391]
The response to this query is 1 * 10 g/mole = 10.
8 0
2 months ago
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