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3 years ago

3. What is a caliber (relate it to rockets)

Engineering

1 answer:

Kamila [148]3 years ago

6 0

Answer:

In this context a caliber is defined as the diameter of the body tube, and it is used to support the general rule of thumb that for a rocket of typical aspect ratio to be stable the CG should be one caliber ahead of CP.

Explanation:

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Can someone draw a electric circuit with electrical symbols for this

erastovalidia [21]

Answer:

i cant draw it

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Explanation:

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3 years ago

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It is easy to say that an organization should hire, reward, and dismiss employees based on their character as well as their know

Nata [24]

It is possible to generate a policy in which common points such as those mentioned above are agreed in order to hire or fire employees in their function of their psychological personality, that is, the character of knowledge and skills. Depending on the company, Test could be created in order to evaluate the psychological skills of the employees, as well as Test to periodically determine how their employees are kept up to date with regard to knowledge. The cumulative filter could be done every semester, for which each employee must exceed a minimum margin of score on these tests, otherwise his position could be at risk.

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This policy agreement is outside the vision and mission of the company, and whose information must be given to the worker once he begins his work activities.

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3 years ago

Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros

umka2103 [35]

Answer:

11.2mm or 0.45in

Explanation:

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3 years ago

Explain why the scenario below describes a team of mechanical engineers.

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Explanation:

because mechanical engineers fix air conditioners cars and other stuffs

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3 years ago

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For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or

Alborosie

Answer:

Glass: Low-Loss dielectric

α = 8.42*10^-11 Np/m

β = 468.3 rad/m

λ = 1.34 cm

up = 1.34*10^8 m/s

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = 9.75 Np/m

β = 12.16 rad/m

λ = 51.69 cm

up = 0.52*10^8 m/s

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = 6.3*10^-4 Np/m

β = 6.3*10^-4 Np/m

λ = 10 km

up = 0.1*10^8 m/s

ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

σ / w*εr*εo

Where, w is the angular speed of wave

εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

$Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\$

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

Glass: Low-Loss dielectric

Tissue: Quasi-Conductor

Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

Glass: Low-Loss dielectric

α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

α = 8.42*10^-11 Np/m

β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

β = 468.3 rad/m

λ = 2*pi / β = 2*pi / 468.3

λ = 1.34 cm

up = λ*f = 0.0134*10^10

up = 1.34*10^8 m/s

ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

α = 9.75 Np/m

β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

β = 12.16 rad/m

λ = 2*pi / β = 2*pi / 12.16

λ = 51.69 cm

up = λ*f = 0.5169*100*10^6

up = 0.52*10^8 m/s

ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

β = α = 6.3*10^-4 Np/m

λ = 2*pi / β = 2*pi / 6.3*10^-4

λ = 10 km

up = λ*f = 10,000*1*10^3

up = 0.1*10^8 m/s

ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

ηc = 6.28*( 1 + j )

8 0

3 years ago