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sleet_krkn [62]
3 years ago
6

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume of 10.0 L, and a temperatu

re of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes, and temperatures as you fill in the following table. P V T A 1 400 kPa 10.0 L 720 K B C 24 L D 15 L (b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A ? B, B ? C, C ? D, and D ? A. Process Q (kJ) W (kJ) ?Eint (kJ) A ? B B ? C C ? D D ? A (c) Calculate the efficiency Wnet / |Qh|. % (d) Show that the efficiency is equal to 1 ? TC / TA, the Carnot efficiency. (Do this on paper. Your instructor may ask you to turn in this work.)

Physics
1 answer:
MrRa [10]3 years ago
7 0

Answer:

Find attached the solution

Explanation:

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A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz
Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, \bar f is given by

\bar f=f\frac {V}{V-V_s} where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and V_s is the speed of car which is unknown, \bar f is given as 440 Hz

440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s

8 0
3 years ago
What is the kinetic energy of an automobile with a mass of 1252 kg traveling at a speed of 12 m/s?
torisob [31]
KE = 1/ 2 * 1252 * 144
 as  KE = 1/2 * m * v ^2
 = 90144 J



4 0
3 years ago
Energy of position or location is called
Orlov [11]
The answer is B!
Explanation: Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.
5 0
3 years ago
What is a derived physical quantity? Name three derived physical quantities, and for each, give its S.I. units and its U.S. Cust
anastassius [24]

Answer:

Physical quantity is a physical property of an object or material that can be expressed by magnitude and unit.

The derived physical quantities are the type of physical quantities which can be expressed or defined by other physical quantities, called the base quantities. Example: Area, Volume, Velocity

Area- SI Unit: m², U.S. Customary unit: acre

Volume- SI Unit: m³, U.S. Customary unit: cubic inch

Velocity- SI Unit: m/s, U.S. Customary unit: ft/s

6 0
3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
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