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suter [353]
2 years ago
13

A vertical wire carries current in the upward direction. An electron is traveling parallel to the wire. What is the angle ααalph

a between the velocity of the electron and the magnetic field of the wire? Express your answer in degrees. View Available Hint(s) ααalpha = nothing degreesdegrees
Physics
1 answer:
Katen [24]2 years ago
5 0

Answer:

First of all note that The magnetic field produced by the vertical wire will be into on the right hand side and it will be out of the page on the left hand side

Assuming that the electron beam is coming from the right hand side of the page parallel to the wire, The direction of the velocity vector(V) is left and the direction of magnetic field due the wire(B) is into the page. If u use right hand rule, you will get the direction downwards but as the formula also depends on q , the charge on electron is negative .Therefore the direction will be inverted i.e Upwards.

If you assume the electron beam coming from left hand side.Then also u will get the same answer.

So, the angle α between the velocity of the electron and the magnetic field of the wire is 90°.

Explanation:

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Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme
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Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider  T_{1}\ and\ T_{2} are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

Let\ Q_{1}\ and\ Q_{2} are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

                                                W\ =\ Q_{1}-Q_{2}

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

                     Efficiency\ [\eta]=\frac{W}{Q_{1}}

                                        \eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}

                                               =\ 1-\frac{Q_{2}} {Q_{1}}

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

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refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a
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Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0
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