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Maurinko [17]
2 years ago
14

What is the last item on the vehicle starting checklist

Engineering
1 answer:
larisa86 [58]2 years ago
4 0

The last item on the vehicle starting checklist before a driver begins to travel is for the driver to release the service brake.

<h3>What does it mean to release the brake?</h3>

The Release Parking Brake message implies that your parking brake is said to be currently engaged and as such one needs to  release the parking brake before one can be able to drive the vehicle.

Therefore, The last item on the vehicle starting checklist before a driver begins to travel is for the driver to release the service brake.

See full question below

what is the last item on the vehicle starting checklist before a driver begins to travel?

Learn more about checklist from

brainly.com/question/16041035

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The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s) in a st
chubhunter [2.5K]

Answer:

\mathbf{t_f = 1436.96 \ sec }

Explanation:

Given that :

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)

i.e

k = 0.14 W/mK

∝ = 6.35 × 10⁻⁸ m²/s

L = 0.01 m

B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\  B_1 = 14.2857

We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number F_o > 0.2

⇒  \dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)

From Table 5.1 ; at B_1 = 14.2857

C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad

\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}]

In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}

-1.9399=-0.001350 *t_f

t_f = \dfrac{-1.9399}{-0.001350}

\mathbf{t_f = 1436.96 \ sec }

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