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Nata [24]
3 years ago
14

A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of

the surroundings). Given these two temperatures, what is the maximum possible efficiency the car can have? (Note that actual car engine efficiencies are in the 20-25% range.) _______ %.
Physics
1 answer:
Step2247 [10]3 years ago
4 0

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

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

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

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

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of
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Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>
  • We have the expression for the initial velocity as,

                           v=\sqrt{2gh}

  • Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

                       g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132

  • Now, the velocity will become,

                        v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s

<h3>How to find the speed of the satellite?</h3>
  • As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

                       v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

brainly.com/question/28108487

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2 years ago
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Here is the answer to the question. Every time when there is a full moon, Mrs. Cook insists that students in her classes display strange behavior and the best way for her to prove the validity of her theory is to r<span>esearch data to find concrete evidence of a link between the moon cycle and human behavior. Hope this helps.</span>
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3 years ago
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PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat
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Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of  1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
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1 + 0.0038T = 14
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3 years ago
A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference
EastWind [94]

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference  ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J

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When a gun is fired at the shooting range, the gun recoils (moves backward). Explain this using the law of conservation of momen
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The total momentum is unchanged according to the law of conservation of momentum. When the gun is fired, the bullet gains a high velocity forward (positive velocity), and that velocity multiplied by its mass is the momentum the bullet gains. Therefore, the gun must gain a momentum backwards to cancel out that momentum forward, so the gun recoils back with a negative velocity.
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