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2 years ago

During an isothermal process, 10 j of heat is removed from an ideal gas. What is the work done by the gas in the process?

Physics

1 answer:

Black_prince [1.1K]2 years ago

8 0

The work done in the isothermal process is 10 joule.

We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.

ΔU = 0

Hence,

ΔU = Q - W

0 = Q - W

Q = W

It means that the heat transferred is the same as the work done.

From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.

Find more on isothermal at: brainly.com/question/17097259

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A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago

The distance between two corresponding parts of a wave is its

oksano4ka [1.4K]

Answer:

wavelenght

Explanation:

The wavelength is the spatial period of a wave, analogous to the temporal period, it is the distance between two consecutive points with maximum amplitude that are repeated in space . In the waves of the sea, the wavelength is easily observed in the separation between two consecutive ridges.

4 0

3 years ago

Read 2 more answers

57. Example of the law of force and acceleration

7nadin3 [17]

Answer:

Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.

Explanation:

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3 years ago

wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h

Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

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3 years ago

Which layer is made of flowing solid rock

ASHA 777 [7]

The mantle is correct

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3 years ago