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 functions specifically designed to be used in conjunction with the insertion (<<) and extraction (>>) operators on stream objects in C++ programming while the 'cout' statement is used to display the output of a C++to the standard output device.
setw: used to specify the minimum number of character positions on the output field
setprecision: Sets the decimal precision to be used to format floating-point values on output operations.
fixed: is used to set the floatfield format flag for the specified str stream.
left: adjust output to the left.
A) To display the number 34.789 in a field of eight spaces with two decimal places of precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;
B) To display the number 7.0 in a field of six spaces with three decimal places of precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;
C) To print out the number 5.789e+12 in fixed-point notation. cout<<fixed<<5.789E12;
(D) To display the number 67 left-justified in a field of six spaces. cout<<left<<setw(7)<<67;
Answer:
2.27secs
Explanation:
The car will take 2.27secs to reach a speed of 10 ft/s.
Kindly go through the attached files for a step by step solution that will show you how the answer was gotten and the normal reaction of both the front and rear wheels.
BOY IS IN THE HOOD BOY IS IN THE HOOD
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Explanation:
Answer:
flow ( m ) = 4.852 kg/s
Explanation:
Given:
- Inlet of Turbine
P_1 = 10 MPa
T_1 = 500 C
- Outlet of Turbine
P_2 = 10 KPa
x = 0.9
- Power output of Turbine W_out = 5 MW
Find:
Determine the mass ow rate required
Solution:
- Use steam Table A.4 to determine specific enthalpy for inlet conditions:
P_1 = 10 MPa
T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg
- Use steam Table A.6 to determine specific enthalpy for outlet conditions:
P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg
x = 0.9 -------------> h_fg = 2392.1 KJ/kg
h_2 = h_f + x*h_fg
h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg
- The work produced by the turbine W_out is given by first Law of thermodynamics:
W_out = flow(m) * ( h_1 - h_2 )
flow ( m ) = W_out / ( h_1 - h_2 )
- Plug in values:
flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )
flow ( m ) = 4.852 kg/s
