In addition to bandages and gauze, it's recommended that a first aid kit also contain

The best saw for cutting miter joints is the

ZanzabumX [31]

Answer:

The best saw for cutting miter joints is the backsaw.

Add-on:

i hope this helped at all.

6 0

3 years ago

(a) Differentiate between heat treatment of ferrous and non-ferrous alloys (b) With your understanding of material's thermal pro

liubo4ka [24]

Answer:

In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.

Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are the alloy element of iron ans carbon.Wrought iron is the purest from of iron.

Heat treatment process for ferrous materials :

1.Normalizing

2.Annealing

3.Quenching

4.Surface hardening

Heat treatment process for non ferrous materials :

Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.

When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.

The use of bimetallic structure -In clock ,thermometers ,engines.

7 0

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:

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

4 0

3 years ago

A wind turbine designer is considering installing a horizontal axis wind turbine at a location in Michigan. To increase the powe

bogdanovich [222]

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P = n * 1/2 * p *Av^3

where: n = turbine efficiency

p = air density

A = πd^2 / 4

v = speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

3 0

2 years ago

Write a loop to print all elements in hourly_temperature. Separate elements with a -> surrounded by spaces. Sample output for

ELEN [110]

The answer & explanation for this question is given in the attachment below.

6 0

3 years ago