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2 years ago

What are the 2 major organizations that develop symbols for use throughout different engineering disciplines

1 answer:

3 0

The disciplines of engineering can be divided into four main categories, chemical, civil, electrical and mechanical engineering. Each main discipline will provide you with a “taste” of the various skills and knowledge required to work in any field related to the discipline.

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A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.

kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current $I = 1$ A

Voltage $V = 12$ V

For finding the resistance,

$V = IR$

$R = \frac{V}{I}$

$R = \frac{12}{1}$

$R =$ 12Ω

(B)

For finding power delivered,

$P = I^{2} R$

$P = (1) ^{2} \times 12$

$P = 12$ Watt

(C)

For finding the potential difference,

$V = IR$

$V = 5 \times 10^{-3} \times 2$

$V = 10 \times 10^{-3}$

$V = 0.01$ V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

ONE sound wave travels through two containers of different gasses. Wave through container 1 has a wavelength of 1.2 m. Wave thro

Zarrin [17]

I’m not 100% but the answer is b

7 0

2 years ago

A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of

kobusy [5.1K]

Answer:

8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.

Explanation:

The total charge accumulated in 1.8 hours will be:

Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)

Total Charge = - 12960 nC = - 12.96 x 10^(-6) C

Since, the charge on one electron is e = - 1.6 x 10^(-19) C

Therefore, no. of electrons will be:

No. of electrons = Total Charge/Charge on one electron

No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]

<u>No. of electrons = 8.1 x 10^13 electrons</u>

6 0

3 years ago

An atom of arsenic has how many electron-containing orbitals

irina [24]

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

$1s^{2}$

$2s^{2}$

$2p^{6}$

$3s^{2}$

$3p^{6}$

$3d^{10}$

$4s^{2}$

$4p^{3}$

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

8 0

3 years ago