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3 years ago

If steam enters a turbine at 600°K and is exhausted at 200°K, calculate the efficiency of the engine.

2 answers:

5 0

Answer: e = 3.333% or 33 1/3%
^
repeated

Identify the given information.
Work input (Wi)= 600 K
Work output (Wo) = 200 K

Choose which formula you should use. In this case, you are finding efficiency.
e = (Wo/Wi) x 100%

Substitute and solve.
e = (200 K/600 K) X 100%
e = 0.333 X 100%
e = 33.333% or 33 1/3%

TiliK225 [7]3 years ago

5 0

Answer: If in you machine you put some heat X, the machine will return some work W and some heat Y. In this case, X = 600k and Y = 200k.

The efficiency of a machine is the ratio of heat transformed into work, so

E = W/X = 1 - Y/X = 1 - 200/600 = 1 - 1/3 = 0.66.

So the efficiency of the turbine is of the 66%

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A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to t

elixir [45]

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by $2\pi r=3.5$

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to $=\frac{3.5}{2}=1.75km$

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

8 0

3 years ago

A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the

insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

$\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}$

Req =3.03Ω , R1 =12.18Ω

$\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}$

$\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}$

$\frac{1}{R2}=0.2479$

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0

4 years ago

determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.

bixtya [17]

Answer:

$F=8.45\times 10^{-8}\ N$

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

$F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N$

So, the force between them is $8.45\times 10^{-8}\ N$ .

8 0

3 years ago

When two substances are mixed together, a color change is observed. Does this mean that a chemical reaction has definitely occur

Elanso [62]

Nope, color change can also occur during a physical change.

7 0

3 years ago

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

5 0

3 years ago