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3 years ago

When turning left, which lane of the intersecting street should you turn onto?

Engineering

2 answers:

nydimaria [60]3 years ago

7 0

You turn onto the left lane

enyata [817]3 years ago

4 0

Answer and Explanation:

The answer is the left lane on the far left of the street you our turning onto, when doing a left turn into it.

This is done for if someone on the road is turning right onto the same road you are turning to, in which they would take the right lane on the far right.

| | |

__| H | |__

L <-

__ ___

D ->

__| | X |___

| | |

X represents the position you are currently on.

D represents the lane with traffic going forward in front of you towards your right.

L represents the lane you need to turn to (the farthest left lane.

R represents the farthest right lane, in which car H would use to turn to.

H represents a car that would turn to position R (right lane).

I hope this helps!

#teamtrrees #PAW (Plant And Water)

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Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

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3 years ago

The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc

Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

$\theta$ = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑ $f_{y}$ = 0, ∑M = 0

If ∑ $M_{A}$ = 0

⇒ $F_{BC}$ ×0.9 - P × 0.6 = 0

⇒ $F_{BC}$ ×3 - P × 2 = 0

⇒ $F_{BC}$ = $\frac{2P}{3}$

If ∑ $M_{B}$ = 0

⇒ $F_{AD}$ ×0.9 = P × 0.3

⇒ $F_{AD}$ ×3 = P

⇒ $F_{AD}$ = $\frac{P}{3}$

Now,

Area, A = $\frac{\pi }{4} X (0.012)^{2}$ = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , $\delta$ = $\frac{P l}{A E}$

Now,

$\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 9.1626 × 10⁻⁹ P

$\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 1.83253 × 10⁻⁸ P

Given that,

$\theta$ = 0.015°

⇒ $\theta$ = 2.618 × 10⁻⁴ rad

So,

$\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}$

⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

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