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3 years ago

What are the three most common metals used in die casting?

Engineering

1 answer:

Anit [1.1K]3 years ago

3 0

Answer:

Aluminium,Copper,Magnesium

Explanation:

The three most common metal of die casting are as follows

1.Aluminium

2.Copper

3.Magnesium

Die casting is the process in which metal is forced in the die to produces the desired casting product.Generally two type of machines are used like cold chamber and hot chamber machining,it depends on the metals.Die casting produces simple shape of casting ,it can not use for complex casting.

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Free ideas free points. You will be reported for answering "no" or I don't know

KengaRu [80]

Answer:

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3 years ago

The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit

zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

$T = \pi \sqrt{\frac{l}{g} }$

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

$T = M^{a}l^{b}g^{c}$

$T = KM^{a}l^{b}g^{c}$

where k is the constant of proportionality.

Now putting the dimensional formula:

$T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}$

$M^{0}L^{0}T^{1} = KM^{a}L^{b+c}$

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

$T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }$

T = $2\pi \sqrt{\frac{l}{g} }$

where k = $2\pi$

8 0

3 years ago

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

2 years ago

A common process for increasing the moisture content of air is to bubble it through a column of water. The air bubbles are assum

likoan [24]

Answer:

Explanation:

Assumptions is that

1. The flow is an unsteady one

2. Bubbles diameter is constant

3. The bubble velocity is slow

4. There is no homogenous reaction

5. It has a one dimensional flux model along the radial direction

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3 years ago

How does the two-stroke Otto cycle differ from the four-stroke Otto cycle?

Digiron [165]

Answer:

Two stroke cycle Four stroke cycle

1.Have on power stroke in one revolution. 1.have one power

stroke in two revolution

2.Complete the cycle in 2 stroke 2.Complete the cycle in 4 stroke

3.It have ports 3.It have vales

4.Greater requirement of cooling 4.Lesser requirement of cooling

5.Less thermal efficiency 5.High thermal efficiency

6.Less volumetric efficiency 6.High volumetric efficiency

7.Size of flywheel is less. 7.Size of flywheel is more.

3 0

2 years ago