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Elena L [17]
3 years ago
6

What are the different types of documents used to communicate engineering designs?

Engineering
1 answer:
Ipatiy [6.2K]3 years ago
7 0

Answer:

COMMON ENGINEERING DOCUMENTS

Inspection or trip reports.

Research, laboratory, and field reports.

Specifications.

Proposals.

Progress reports.

ect...

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
The thermal efficiency of two reversible power cycles operating between the same thermal reservoirs will a)- depend on the mecha
mestny [16]
C ,, i’m pretty sure .
4 0
4 years ago
Please write the following code in Python 3. Also please show all output(s) and share your code.
maksim [4K]

Answer:

sum2 = 0

counter = 0

lst = [65, 78, 21, 33]

while counter < len(lst):

   sum2 = sum2 + lst[counter]

   counter += 1

Explanation:

The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>

While there are numbers in the <em>lst</em>,  loop through <em>lst</em>

Add the numbers in <em>lst</em> to the sum2

Increment <em>counter</em> by 1 after each iteration

6 0
3 years ago
What unit of measurment would be used to measure current?
Alex_Xolod [135]

Answer:

The S. I unit of current is Amphere

5 0
3 years ago
How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work
lozanna [386]

Answer:

its so simple. u must mind some formulas.

Explanation:

mili->10^(-3)

micro->10^(-6)

nano->10^(-9)

so write the exact number and move "." to left or right depend question.

in this one:

23.5 is 23500000.0 nano with a default dot at the end

for turning to mili u must move the dot 6 steps to left so it will be: 23.5 milimeter.

6 0
3 years ago
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