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Murljashka [212]
3 years ago
12

The cycle used by the engine is composed of three processes: a process at constant pressure between point 1 and 2, a process at

constant volume between points 2 and 3, and an isothermal process between points 3 and 1. What are the processes between points 1 and 2 and between points 2 and 3, respectively
Physics
1 answer:
aev [14]3 years ago
5 0

Answer:

Isobaric process between points 1 and 2 while isochoric process between 2 and 3.

Explanation:

A themodynamic equilibrium system which changes from initial state to the final state by a process is known as thermodynamic processes. There are many different kinds of thermodynamic process which depends upon the thermodynamic quantities such as pressure, volume, temperature, entropy etc. The most common thermodynamic processes are isobaric, isothermal, isochoric and adiabatic.

When the temperature of the system remains same during the thermodynamic process, the process is known as isothermal.

When the heat of the system remains same during the thermodynamic process, the process is known as adiabatic.

When the volume of the system remains same during the thermodynamic process, the process is known as isochoric.

When the pressure of the system remains same during the thermodynamic process, the process is known as isobaric.

According to the question, between points 1 and 2 pressure remains constant, hence the process is isobaric.

According to the question, between points 2 and 3 volume remains constant, hence the process is isochoric.

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8 0
3 years ago
The current through a 0.2-H inductor is i(t) = 10te–5t A. What is the energy stored in the inductor?
lakkis [162]

Answer:

E = 10t^2e^-10t Joules

Explanation:

Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.

The energy E stored in the inductor can be expressed as

E = 1/2Ll^2

Substitutes the inductor L and the current I into the formula

E = 1/2 × 0.2 × ( 10te^-5t )^2

E = 0.1 × 100t^2e^-10t

E = 10t^2e^-10t Joules

Therefore, the energy stored in the inductor is 10t^2e^-10t Joules

6 0
4 years ago
A student performs an experiment and must measure the lengths of four different objects: a textbook, a pencil, a cup, and a piec
Strike441 [17]
Most appropriate unit of measurement should be centimetres, cm.
7 0
4 years ago
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Oxygen has 6 valence electrons. how many hydrogen atoms (who have 1 valence electrons) can bond with it? remember the octet rule
astra-53 [7]

Answer:

Oxygen starts with six valance electrons and ends with six valance electrons but after bonding has parts of the densities of 8 electrons

Explanation:

Chemical bonding does not increase the valance electrons. Oxygen has 6 valence electrons. Bonding is sharing of valance electrons

The electron density of two of the valance electrons of Oxygen are shared with the Hydrogen atoms. The Hydrogen atoms is turn share part of the electron density of their single electron with Oxygen.

This means that there are parts of 8 electron densities around the Oxygen. This does not increase the number of valance electrons of Oxygen. (Note the sharing is very unfair, Oxygen gets more than its

share of the electron densities of the two electrons used for the chemical bond.

Explanation:

5 0
3 years ago
A cable that weighs 4 lbs/ft is used to lift a sandbag up a mine shaft 50 feet deep. A small tear in the bag causes sand to leak
sveticcg [70]

Answer:

W = 7500 lbs/ft

Explanation:

First, the question is incomplete. The missing part is How much work was done by lifting the bag this far?

Now that we need to calculate the work done to pull the sandbag, the best way to do this is assuming a linear function for the weight of the sandbag as it reach the height of the mine.

Now, the expression to calculate the work is:

W = W1(lift cable) + W2(lift sandbag)

To get work to lift the cable:

W = ∫2xdx

W1 = ∫2(50)dx = 50² = 2500 lbs/ft

To get the work to lift the bag, we should remember that we are going to assume that this is a linear function, so, y = ax + b

The work:

W = ∫F dx (1)

F = ax + b

b is the final weight of the bag, which is 60.

Solving for a:

a = m2 - m1 / x2 - x1

m1 and m2 are the mass of the sandbag, and x1 and x2 the distance of the sandbag.

We have this data already, so:

a = 80 - 60 / 50 - 0 = 0.4 or 2/5

F = 2/5x + 90 (2)

Replacing in (1)

W2 = ∫2/5x + 90 dx

W2 = ∫2/5x dx + ∫90 dx

W2 = 2/5 ∫x dx + 90 ∫dx

W2 = 2/5 x²/2 + 90∫dx

W2 = x²/5 + 90 x (from 0 to 50)

W2 = 50² / 5 + 90*50

W2 = 5000 lbs/ft

Finally the work done is:

W = W1 + W2

W = 2500 + 5000 = 7500 lbs/ft

4 0
3 years ago
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