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3 years ago

7

A carnot engine operates between two heat reservoirs at temperatures th and tc. an inventor proposes to increase the efficiency

by running one engine between th and an intermediate temperature t and a second engine between t and tc using as input the heat expelled by the first engine.

1 answer:

Sonja [21]3 years ago

5 0

The efficiency of the first Carnot engine is
n1 = 1 - Th/T
The efficiency of the second Carnot engine is
n2 = 1 - T/Tc
The total efficiency of the engines put in series is
n = 1 - Th/Tc
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In general, what do you think are the benefits if you already achieve your body goals.

balandron [24]

Answer:

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Explanation:

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3 years ago

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

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Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee

musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

$Speed = \frac{Distance}{Time}$

$\therefore 343 \ m/s= \frac{Distance}{5 \, s}$

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is $Time = \frac{Distance}{Speed} = \frac{1715}{299,792,000} = 0.00000572 \, s$ which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0

3 years ago

What sphere on earth includes mountains

Lostsunrise [7]

I believe it is lithosphere

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3 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

3 years ago