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krok68 [10]
2 years ago
5

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

reservoir. during that time, what is the maximum amount of work wmax that the engine might have performed?
Physics
1 answer:
tino4ka555 [31]2 years ago
6 0

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
horrorfan [7]

Answer:

The value is t_1 =  9 \  s

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

v  =  u   +  at

Here u = 0 \ m/s

a = 13.5 m/s^2

So

v  =  0   +  13.5 *  3.10

=> v  =  41.85 \ m/s

The is distance it covers at this time is

s =  u *  t  +  \frac{1}{2} a *  t^2

=> s =  +  \frac{1}{2} * 13.5 *  3.10^2

=> s =64.87

Now when sled stops its the final velocity is v_f =  0 m/s while the initial velocity will be the velocity after its acceleration i.e v  =  41.85 \ m/s

So

v_f  =  v  +  a_1t_1

Here  a_1 =  - 4.65, the negative sign shows that it is deceleration

So

           0  =  41.85  - 4.65 *  t_1

=> t_1 =  9 \  s

3 0
2 years ago
Calculate the force of gravity between two objects of masses 1300 kg and 7800 kg, which are 0.23 m apart.
uranmaximum [27]

Answer:

F = Gm1m2/r^2 where G = 6.67x10^-11, m1 =1300, m2 = 7800, r = 0.23m

F = 6.67x10^-11 *1300*7800/(0.23)^2 = 0.0127852N

Explanation:

6 0
2 years ago
A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other
san4es73 [151]

Answer:

16,18,22

Or

1,3,7

Explanation:

The detailed explanation is contained in the image attached. The lengths are found using Pythagoras theorem and the two lengths reflects the two values of x yielded by the quadratic equation

8 0
2 years ago
Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una
EleoNora [17]

Answer:

a)  f = 615.2 Hz      b)  f = 307.6 Hz

Explanation:

The speed in a wave on a string is

         v = √ T / μ

also the speed a wave must meet the relationship

          v = λ f

           

Let's use these expressions in our problem, for the initial conditions

            v = √ T₀ /μ

             √ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

         T = 2T₀

          √ (T /μ) = λ f

          √( 2To /μ) = λ f

         √2  √ T₀ /μ = λ f

we substitute

         √2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

           λ₀ = λ

           f = f₀ √2

           f = 435 √ 2

           f = 615.2 Hz

b) The tension is cut in half

         T = T₀ / 2

         √ (T₀ / 2muy) =  f = λ f

          √ (T₀ / μ)  1 /√2 = λ f

           fo / √2 = f

           f = 435 / √2

           f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

         v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

          v= λ f  

           

Usemos estas expresión en nuestro problema, para las condiciones iniciales

            v= √ To/μ

             √ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

         T = 2T₀

          √ ( T/μ) = λf

          √ ) 2T₀/μ = λ f

         √ 2 √ T₀/μ = λ f

         

substituimos  

         √2    ( λ₀ f₀)  =  λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

                 λ₀ =  λ

           f = f₀ √2

           f = 435 √2

           f = 615,2 Hz

b)  La tension se reduce a la mitad

         T = T₀/2    

         RA ( T₀/2μ)  =  λ  f

          Ra(T₀/μ) 1/ra 2  =  λ f

           fo /√ 2 = f

           f = 435/√2

           f = 307,6 Hz

5 0
2 years ago
Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The mass o
insens350 [35]

Answer:

295.42 N

Explanation:

From Newton's law of universal gravitation.

F = Gmm'/r².................. Equation 1

Where F = Gravitational force, G = Universal constant, m = mass of the human, m' = mass of mass, r = radius of mass.

Given: m = 80 kg, m' = 6.4×10²³ kg, r = 3.4×10⁶ m.

Constant: G = 6.67×10⁻¹¹ Nm²/Kg²

Substitute into equation 1

F =  6.67×10⁻¹¹(80)(6.4×10²³ )/( 3.4×10⁶)²

F = 3415.04×10¹²/(11.56×10¹²)

F = 3415.04/11.56

F = 295.42 N

Hence the gravitational force =  295.42 N

5 0
2 years ago
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