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2 years ago

Technician A says that the airbag computer retains power in case the power is lost after an initial collision.

Engineering

1 answer:

earnstyle [38]2 years ago

8 0

Answer:

A. A only

Explanation:

Airbags are safety features added to vehicles to prevent or reduce the shock impact during a collision.

Disconnecting the negative battery cable would prevent power from reaching the airbag system, but the airbag system has a capacitor that stores charges at initial incidents and would deploy the bags. After a collision, the airbag leads are shot off preventing the airbag system power.

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An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.

saul85 [17]

Answer:

a. $L_o$ = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, $L_o$ = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

For the school, $K_{LL}$ = 2

Therefore, we have;

$L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf$

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, $W_d$ = b × $W_{D + L}$ =

∴ $W_d$ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, $W_d$ = 812.8 ft./lb

d. For the uniformly distributed load, we have;

$V_{max}$ = 812.8 × 26/2 = 10566.4 lbs

$M_{max}$ = 812.8 × 26²/8 = 68,681.6 ft-lbs

$v_{max}$ = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

$P_L$ = 1000 lb

$W_{D}$ = 20 × 16 = 320 lb/ft.

$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0

2 years ago

You are installing network cabling and require a cable solution that provides the best resistance to EMI.Which of the following

gregori [183]

Answer: b. STP

Explanation:

Twisted Pair Cables are best used for network cabling but are usually prone to EMI (Electromagnetic Interference) which affects the electrical circuit negatively.

The best way to negate this effect is to use Shielding which will help the cable continue to function normally. This is where the Shielded Twisted Pair (STP) cable comes in.

As the name implies, it comes with a shield and that shield is made out of metal which can enable it conduct the Electromagnetic Interference to the ground. The Shielding however makes it more expensive and in need of more care during installation.

7 0

3 years ago

What action below would tell your computer to "Send" an email?

Fynjy0 [20]

Click I think:) not sure tho

4 0

2 years ago

Read 2 more answers

A programmer needs to understand sequencing to determine whether the order of steps will affect the outcome of the program.

Ivahew [28]

I would say true as a programmer must know the sequence to understand the steps and they must also know the outcome of the program(that’s the reason they are programming)

7 0

2 years ago

Read 2 more answers

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 37 MPam. It has been de

Harman [31]

Answer:

we are supposed to determine the stress level at which a wing component on an aircraft will fracture for a given toughness of $37 MPa\sqrt{m}$ and maximum internal crack length $1.29 mm$ , given that fracture

occurs for the same component using the same alloy at one stress level $206 Mpa$ and another internal crack length (2.58 mm). It first becomes necessary to solve for the parameter Y for the conditions under which fracture occurred , therefore

$Y=\frac{K_{lc} }{sigma\sqrt{\pi a } }$

$=(37Mpa\sqrt{m} )/(206Mpa)\sqrt{\pi \frac{2.58*10^-^3m}{2} } \\=2.82$

$sigma_{c} =K_{lc} /Y\sqrt{\pi a }$

$=(37MPa\sqrt{m} )/(2.82)\sqrt{\pi \frac{1.29*10^-^3}{2} } \\= 293 MPa$

5 0

2 years ago