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Annette [7]
3 years ago
14

An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, conver

ts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R. Are these measurements reasonable? Why?
Engineering
1 answer:
lesya [120]3 years ago
3 0

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

You might be interested in
The advantages of solar cells include all of the following, except a.moderate net energy yield b.little or no direct emissions o
Xelga [282]

Answer:

C

Explanation:

One of the disadvantages of solar cells is that electricity storage systems are not readily available. Excess energy generated by the solar panels are wasted except they are stored by solar batteries for later use. There are various systems for storing electricity from solar cells apart from solar batteries which is the common storage system. An example of another electricity storage system for solar cell is using the water electrolyzer to store solar energy which can be used to later generate hydroelectricity.

Advantages of a solar cell includes Renewable energy, Economy-friendly and environmental-friendly energy and good durability

6 0
3 years ago
Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x
RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

>>

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

>>

Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 0
3 years ago
Explain 3 ways that people in sports use engineering to increase their performance?
LenKa [72]
Designing systems for manufacturing, motion analysis or impact testing;
building and testing prototypes;
analyzing the human body to prevent injury;
developing or designing new light weight materials that will be more comfortable and withstand greater impacts or forces;
7 0
3 years ago
1.0•10^-10 standard form
Drupady [299]

Answer:

1.0 * 10^{-10} = 0.0000000001

Explanation:

Given

1.0 * 10^{-10}

Required

Convert to standard form

1.0 * 10^{-10}

From laws of indices

a^{-x} = \frac{1}{a^x}

So, 1.0 * 10^{-10} is equivalent to

1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}

1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}

1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}

1.0 * 10^{-10} = 1.0 * 0.0000000001

1.0 * 10^{-10} = 0.0000000001

Hence, the standard form of 1.0 * 10^{-10} is 0.0000000001

3 0
3 years ago
A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl
german

Answer:

BOD concentration at the outflow = 17.83 mg/L

Explanation:

given data

flow rate of Q = 4,000 m³/day

BOD1 concentration of Cin = 25 mg/L

volume of the pond = 20,000 m³

first-order rate constant equal = 0.25/day

to find out

What is the BOD concentration at the outflow of the pond

solution

first we find the detention time that is

detention time t = \frac{volume}{flow rate}

detention time t = \frac{20000}{4000}

detention time = 5 days

so

BOD concentration at the outflow of pond is express as

BOD concentration at the outflow = Cin ( 1 - e^{-kt} )

here k is first-order rate constant and t is detention time and Cin is BOD1 concentration

so

BOD concentration at the outflow = 25 ( 1 - e^{-0.25(5)} )

BOD concentration at the outflow = 17.83 mg/L

8 0
3 years ago
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