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daser333 [38]
3 years ago
6

One of the most efficient heat engines ever built is a coal-fired steam turbine in the Ohio River valley, operating between 1 87

0°C and 430°C.
(a) What is its maximum theoretical efficiency? %

(b) The actual efficiency of the engine is 42.0%. How much mechanical power does the engine deliver if it absorbs 1.30 105 J of energy each second from its hot reservoir?
Physics
1 answer:
k0ka [10]3 years ago
5 0

Answer:The answer

Explanation:

You might be interested in
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.
Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0
3 years ago
The temperature of the liquid in a container decreases as the liquid evaporates. Use kinetic theory to explain why.
melisa1 [442]

Answer:

Temperature decreases because the number of collision of the molecules decreases as they escape or evaporate. Molecules are in constant motion. Increase in temperature leads to increase in average kinetic energy of the molecules.

6 0
2 years ago
The center of mass of the system is the point at which the total mass of the system could be concentrated without changing the _
Sveta_85 [38]

  The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

  of the system.

Centre of Mass is the point at which the whole mass of the system

is assumed to be concentrated.

 The general formula for the COM is:

              xₙ =  Σmₐxₐ / Σmₐ         where,  a = 1,2,3.........n

  Here the term Σ mₐ xₐ is called the first moment of the system and the

  denominator expression is called total mass of the system.

    Therefore, from this theory we can say that the moment of the system

    remain unchanged while calculating the COM.

  Hence, The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

  of the system.

   learn more about centre of mass here:

             brainly.com/question/3454419

                #SPJ4

8 0
1 year ago
Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is
Llana [10]

Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

4 0
3 years ago
Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to
andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3   ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3  ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = \frac{m1v}{3v3}  -----  ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 =  2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1  hence v2 = 2v1  where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴  v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0
3 years ago
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