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natima [27]
3 years ago
6

What is digital communication?​

Engineering
1 answer:
mart [117]3 years ago
4 0

Answer:

Any exchange of data that transmits the data in a digital form is a digital communication.

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Seeing what the other side of the world is doing right now
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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 132mm diameter solid circular section​
Ganezh [65]

Answer:

not sure if this helps but

5 0
3 years ago
Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125
Anna [14]

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

4 0
11 months ago
What is (10 to the power of three) times (3 to the power of 10)? will give brainliest
VLD [36.1K]

59,049,000 is the answer.

7 0
3 years ago
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