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2 years ago

7. If you can't ignore a distraction, what should you do?

Engineering

1 answer:

Mice21 [21]2 years ago

3 0

If you can't ignore a distraction, as a driver you should: D. Both A and B.

<h3>What is a distraction?</h3>

A distraction can be defined as any form of event that is capable of making a driver to loose concentration or lure his or her eyes away from the road.

This ultimately implies that, a distraction can cause a driver to experience a road accident (car crash) if not properly managed or ignored.

In conclusion, if you can't ignore a distraction, as a driver you should pull over or take care of it at your next planned stop.

Read more on traffic regulations here: brainly.com/question/22768531

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The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is

Morgarella [4.7K]

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( $d^{2}$ - $d_{0} ^{2}$ = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = $d^{2} - d^{2} _{0}$ = kt

d = final grain size

$d_{0}$ = initial grain size

k = constant ( temperature dependent )

t = 0

8 0

3 years ago

A driver is traveling at 90 km/h down a 3% grade on good, wet pavement. An accident

Paul [167]

Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ - μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction coefficient is 0.35

4 0

3 years ago

A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa

KengaRu [80]

Answer:

stress = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement $\delta = 1.25 \mu m =\frac{1.25}{1000} = 0.00125 mm$

stress due to displacement in structural steel can be determined by using following relation

$E =\frac{stress}{strain}$

$stress = E \times strain$

where E is young's modulus of elasticity

E for steel is 200 GPa

$stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}$

stress = 50MPa

7 0

3 years ago

When we utilize a visualization on paper/screen, that visualization is limited to exploring: Group of answer choices Relationshi

Mila [183]

Answer:

As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

6 0

3 years ago

A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The

AleksandrR [38]

Answer:

The work furnished by the compressor is $69.77kJ/s$

The minimum work required for the state to change is $55.26kW$

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

8 0

3 years ago