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3 years ago

At what times should you use your headlights?

Engineering

2 answers:

SVETLANKA909090 [29]3 years ago

8 0

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

hope this helps

storchak [24]3 years ago

5 0

Answer:

at night time from sunset all the way till sunrise

Explanation:

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Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul

Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

$\mu =1.846 \times 10^{-5} Pa-s$

$\rho =1.77 kg/m^3$

And reynold's number is given by

$Re.=\frac{\rho v\time x}{\mu }$

$5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}$

$x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}$

x=3.47 m

5 0

4 years ago

__________ affect(s) the amount of air conditioning required.

Grace [21]

Answer:

Option D

All the above

Explanation:

Depending with the number of occupants in a building, the number of air conditioners required can either be increased or reduced. For instance, if the building is to be a classroom of over 50 students, 1 air-conditioner can't serve effectively. Similarly, the activity of occupants also dictate the amount of air conditioners required since if it's a gym room where occupants exercise often then the air conditioners required is different from if the room was to serve as a lounge. The appliances that also operate in a room require that air conditioners be installed as per the heat that may be generated by the appliances.

5 0

4 years ago

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

Be-16 a garbage dumping placard must be prominently posted on boats longer than what size?

Crank

Answer:

26 feet and longer boats that have garbage dumping placard must be prominently posted and the boats which are 40 feet and longer must have the written waste management plan.

4 0

3 years ago

4. Which type of duct undergoes more rigorous testing before it's labeled?

Alenkasestr [34]

Answer:

Flexible duct

Explanation:

it is flexible duct due to current flow

7 0

3 years ago