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2 years ago

Two importance ofheadlight alignment

Engineering

1 answer:

Hitman42 [59]2 years ago

8 0

Answer:

Proper horizontal and vertical adjustments.

Explanation:

Proper horizontal and vertical adjustments will ensure that you can see far enough down the road to react to obstacles or avoid animals.

You might be interested in

What's the difference between accuracy and percision in measuring and gaging?

lidiya [134]

Answer:

The term Accuracy means that how close our result to the original result.

Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.

And the term Precision means how likely we get result like this.

Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.

4 0

3 years ago

The viscosity of a fluid is 10 be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diamete

a_sh-v [17]

Answer:

0.023 Pa*s

Explanation:

The surface area of the side of the inner cylinder is:

A = π*d*l

A = π*0.15*0.75 = 0.35 m^2

At 200 rpm the inner cylinder has a tangential speed of:

u = w * r

u = w * d/2

w = 200 rpm * 2π / 60 = 20.9 rad/s

u = 20.9 * 0.15 / 2 = 1.57 m/s

The torque is of 0.8 N*m, this means that the force is:

T = F * r

F = T / r

F = 2*T / d

For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:

F = μ*A*u / y

Where

μ: viscocity

y: separation between pates

A: surface area of the plates

Then:

2*T / d = μ*A*u/y

Rearranging:

μ = 2*T*y / (d*A*u)

μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s

5 0

2 years ago

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

$h=\frac{2\sigma cos(\alpha )}{\rho gd}$

where

$\sigma$ is the surface tension of the liquid

$\alpha$ is the contact angle of the liquid

$\rho$ is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

$\alpha =0$

$\rho _{w}=1000kg/m^3$

Applying the values we get

$h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm$

Part b) When the liquid is mercury:

For mercury and glass we have

$\sigma =485.5\times 10^{-3}N/m$

$\alpha =138^o$

$\rho _{w}=13.6\times 10^{3}kg/m^3$

Applying the values we get

$h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm$

The negative sign indicates that there is depression in mercury in the tube.

4 0

2 years ago

4. In the Hyatt Regency walkway case study, it is reported that Jack Gillum stamps the 42 shop drawings, including the revised S

mario62 [17]

Answer:

Responsibility

Explanation:

By stamping the drawings that he was looking over, Jack Gillum conveys the fact that he is accepting responsibility for this work. The purpose of Gillum's stamp is to explain that such work has been under engineering review, and that it has fulfilled all the requirements that he watches our for. By putting his stamp in this work, Gillum accepts responsibility in case an error or a discrepancy is found in the drawings.

3 0

2 years ago

A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

$W_{f}=\gamma(L)(B)(D)$

Where $W_{f}$ is the weight of footing, γ is the unit weight of concrete, L is the length of footing is the width of footing, and D is the depth of footing

Substitute $2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3}$ for γ in the equation

$\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}$

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

$\sigma_{z p}^{\prime}=\gamma(D+B)-u$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0$ for u in the equation

$\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}$

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

$\sigma_{z D}^{\prime}=\gamma D$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3}\) for $\gamma, 0.5 m$ for $D$ and 0 for \(u$ in the equation.

$\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}$

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

$I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}$

Here $I_{e p}$ is the influence factor at the midpoint of soil layer $\sigma_{z^{p}}^{\prime}$ is initial vertical stress, $\sigma_{z^{p}}^{\prime}$ is vertical effective stress, and $Q$ is bearing pressure

Substitute $36 kPa for $\sigma_{z p}^{\prime}, 228.47$ kPa for $q,$ and 9 kPa for $\sigma_{z D}^{\prime}$$ in the equation $\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}$

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0

3 years ago

Two importance ofheadlight alignment​

Two importance ofheadlight alignment