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Reptile [31]
2 years ago
12

A Carnot engine has an efficiency of 83.0% and performs 4500 J of work every cycle. How much energy is discharged to the lower t

emperature reservoir every cycle?
Physics
1 answer:
ololo11 [35]2 years ago
6 0

We will start by finding the heat in state one through the thermal efficiency or efficiency of a thermal machine is a coefficient or dimensionless ratio calculated as the ratio of the energy produced (in an operating cycle) and the energy supplied to the machine. Mathematically it is the relationship between the work generated and the heat emanated.

So,

\eta = \frac{W}{Q_1} \rightarrow Q_1 = \frac{W}{\eta}

Q_1 = \frac{4500}{0.83}

Q_1 = 5421J

The total change of energy is equivalent to 4500J and this is equal by conservation of energy to the total change in heat. So:

W = \Delta Q

W = Q_1-Q_2

4500 = 5421- Q_2

Q_2 = 921.68 J

Therefore the energy which is discharged to the lower temperature reservoir every cicle is 921.7J

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Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi
Katyanochek1 [597]

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

7 0
2 years ago
Give reasons for the following:
Juliette [100K]

Answer:

a. lower surface area, less resistence

b. more surface area, the load is split so no single tire overstrained

c. more surface area, more resistance against the sand. human steps sink down in the sand.

d. rapid change in air pressure on eardrums lead to somewhat-painful tension

e. air would always find its way in so no pressure difference can be achieved

(would indeed appreciate the brainliest if you appreciate the work)

4 0
3 years ago
Which liquid is the most vicious? Syrup, water, milk, or apple juice?
Nadya [2.5K]

Answer:

syrup

Explanation:

6 0
3 years ago
An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless
Elenna [48]

Answer:

Explanation:

Potential energy lost by mass = mgh

= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

= 1/2 m v² = 1/2 k x² , k is spring constant , x is compression , m is mass falling down

.5 x 10 v² = .5 x 500 x .75²

v = 5.3 m /s

b ) kinetic energy of mass at the bottom

= /2 m v²

= .5 x 10 x 5.3²

= 140.45 J

energy lost by mass while coming down

=potential energy at the top - kinetic energy at bottom

=  196 - 140.45

= 55.55 J .

This is equal to negative work done by friction

work done by friction = - 55.55 J

c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45  - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

h = .866 m

( We have assumed that loss of energy in return journey will be same due to friction . )

6 0
3 years ago
Neutron stars are extremely dense objects formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose th
mihalych1998 [28]

Answer:

10989.55932 rad/s

Explanation:

m = Mass of object

M = Mass of neutron star = 2\times 1.989\times 10^{30}\ kg

R = Radius of neutron star = 13000 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

\omega = Angular speed

Here, the gravitational force will balance the centripetal force

\dfrac{GmM}{R^2}=mR\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{GM}{R^3}}\\\Rightarrow \omega=\sqrt{\dfrac{6.67\times 10^{-11}\times 2\times 1.989\times 10^{30}}{13000^3}}\\\Rightarrow \omega=10989.55932\ rad/s

The greatest possible angular speed an object can have is 10989.55932 rad/s

8 0
3 years ago
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