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shepuryov [24]
3 years ago
15

Three Carnot engines operate between the following temperature limits.

Physics
1 answer:
Kipish [7]3 years ago
6 0

Answer:c>a>b

Explanation:

Given

All the engine extracts same amount of heat(Q) from High-temperature reservoir

For a) 400 and 500 K

\eta _{engine}=1-\frac{T_L}{T_H}

\eta _{engine}=\frac{work\ supplied}{heat\ supplied}

1-\frac{400}{500}=\frac{W_a}{Q}

W_a=\frac{Q}{5}

For b)500 K and 600K

1-\frac{500}{600}=\frac{W_b}{Q}

W-b=\frac{Q}{6}

For c) 400 K and 600 K

1-\frac{400}{600}=\frac{W_c}{Q}

W_c=\frac{2Q}{3}

So c will give the highest amount of work

c>a>b

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xxMikexx [17]
At  a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:

<span>S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M
</span>
At 0.890 atm,
<span>S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M</span>
<span>
If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows</span>
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3.30 L x (0.000273 mol / L) = 0.0012012 mol</span>

3.30 L x (0.00024297 mol / L) = 0.001069068 mol

0.0012012 mol - 0.001069068 mol = 0.000132 mol
6 0
4 years ago
Which term describes the amount of charge that passes a point in a circuit each second?
mestny [16]
Current, I got it right on my quiz
5 0
2 years ago
A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b
Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 m/s^2

So 4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}

Now at time 2t,

  S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g m/s^2

4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}

At time 2t,

     S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm

 So vertical distance traveled in time 2t = 16 cm to the bottom

 

6 0
3 years ago
A laser beam strikes a plane's surface with an angle of52.
Vadim26 [7]

Answer:

Angle between incident ray and reflected ray will be 104°

Explanation:

We have given angle of incidence = 52 °

From law of reflection angle of incidence will be equal to angle of reflection

So angle of reflection will be also 52°

We have to find the angle between incident ray and reflected ray

As the incidence angle and reflected angle both is from normal of the surface and opposite to each other

So angle between incident ray and reflected ray will be 52°+52° = 104°

7 0
4 years ago
A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north un
nexus9112 [7]

Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0
3 years ago
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