An engine has been diagnosed with blowby.

Write cout statements with stream manipulators that perform the following:

Semenov [28]

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;

7 0

4 years ago

This came off my car earlier today. What is it and what does it do?

patriot [66]

Answer:

get a new car or duck tape it back on or go get it fixed

Explanation:

that's just whut I wold do.

6 0

3 years ago

Read 2 more answers

A ball is dropped from rest from the top of a cliff that is 30 m high. From ground

trasher [3.6K]

The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

<u>Given the following data:</u>

Initial velocity = 0 m/s (since the ball is dropped from rest).

Height = 30 meters.

<u>Scientific data:</u>

Acceleration due to gravity (a) = 9.8 $m/s^2$ .

To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.

<h3>How to calculate the velocity.</h3>

Mathematically, the third equation of motion is given by this formula:

$V^2 = U^2 +2aS$

<u>Where:</u>

V is the final velocity.

U is the initial velocity.

a is the acceleration.

S is the distance covered.

Substituting the parameters into the formula, we have;

$V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}$

V = 24.25 m/s.

<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.

The time at which the two balls meet is calculated as:

$Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}$

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

$y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters$

Lastly, the distance below the top of the cliff is calculated as:

$Distance = 30-22.47$

Distance = 7.53 meters.

Read more on distance here: brainly.com/question/10545161

4 0

3 years ago

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap

alexdok [17]

Answer:

heat rate = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

= 5.6*10^-4/3.59*10^-7

= 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

=π * 0.005 * 2

=0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

= 29.34* 0.0314(160-151.99)

= 7.38 W

the Prandtl number using the formula

5 0

3 years ago