Can I have some help please? It would mean a lot!

Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and

Thepotemich [5.8K]

First law of thermodynamics:

It states that "Energy neither be created nor it can be destroyed". simply it converts one form of energy into another form.

It is also known as "law of conservation of energy"

Limitations of First law

It doesn't provide a clear idea about the direction of transfer of heat.
It doesn't provide the information that how much heat energy converted inti work.
Its not given any practical applications.

II law of thermodynamics:

It states that "the total entropy of the system can never decrease over time"

It is strongly proved by two laws, they are

1. Kelvin-plank statement:

He stated that "any engine does not give 100% efficiency". It violates the Perpetual motion of machine II kind(PMM-II).

2. Classius statement: 

  It states that "Heat always flows from high temperature body to low temperature body, without aid of external energy".

Also it stated that " Heat can also be transferred from low temperature body to high temperature body, by the aid of an external energy".

Applications of II law: 

Refrigeration &Air conditioning, Heat transfer, I.C. engines, etc.

3 0

3 years ago

Read 2 more answers

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its loc

MArishka [77]

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

$V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11$

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

$distance=v\,*\, t$

$distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11$

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

6 0

3 years ago

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

7 0

3 years ago

Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of

notsponge [240]

Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force ( $F$ ), measured in newtons, is the product of mass ( $m$ ), measured in kilograms, and net acceleration ( $a$ ), measured in meters per square second. That is:

$F = m\cdot a$ (1)

The initial force applied in the mass is:

$F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)$

$F = 25\,N$

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to $\frac{4}{5}$ of the initial force. The acceleration of the mass is: