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Airida [17]
2 years ago
6

Why is electricity considered a secondary source of energy

Engineering
1 answer:
Law Incorporation [45]2 years ago
6 0

Answer:

The electricity that we use is a secondary energy source because it is produced by converting primary sources of energy such as coal, natural gas, nuclear energy, solar energy, and wind energy, into electrical power.

Explanation:

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About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

7 0
2 years ago
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the mot
Thepotemich [5.8K]

78950W the answer

Explanation:

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

5 0
3 years ago
How does java achieve portable
sergejj [24]

Answer:

Java is called portable because you can compile a java code which will spew out a byte-code, and then you run that code with Java Virtual Machine. Java Virtual Machine is like an interpreter, which reads the compiled byte-code and runs it. So first of all, you need to install the JVM on the system you want.

Explanation:

5 0
3 years ago
You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is
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14256000. Kanjiuijhgg
5 0
3 years ago
Read 2 more answers
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
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