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3 years ago

Airbags will deploy in a head on collision but not in a collision that occurs from angle

Engineering

1 answer:

Aneli [31]3 years ago

8 0

Answer:

Airbags will deploy in almost any angle.

Explanation:

Cars have sensors around them, so when the car gets hit, the sensors detect a crash and deploy the airbags to keep you safe.

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3 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

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3 years ago