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Rudiy27
3 years ago
10

The following is a list of metals and alloys:

Engineering
1 answer:
viktelen [127]3 years ago
5 0

Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

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Thermal energy is...
Yuki888 [10]
B because thermal has to do with temperature and it’s the amount of kinetic and potential energy in and object
4 0
2 years ago
Read 2 more answers
Suppose a large amount of power is required. Which engine would you choose between Otto and Diesel? Why?
Firdavs [7]

Answer:

Otto engine

Explanation:

As we know that

Power = Torque x speed

So we can say that when speed of engine then power of engine also will increases.

The speed of Otto engine is more as compare to Diesel engine so the power of Otto engine is more.But on the other hand torque of Diesel engine is more as compare to Otto engine but the speed is low so the product of speed and torque is more for Otto engine .It means that when requires large amount of power then Otto engine should be use.

6 0
3 years ago
(a) what is Linear equation (b) Why Laplace's equation is linear
Triss [41]

Answer:

 A) Linear Equation -

      Linear equation has only one independent variable and when the linear equation plotted on a graph it forms a straight line. It is made up of two expressions equal to each other in a equation. Linear equation graph fits the Y= mx+a ( m=slope).

B) Laplace's equation is linear as it is a second order partial differential equation. So if we put dependent variable in differential equation it always show result in linear.

7 0
3 years ago
The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical
Nookie1986 [14]

Solution:

Given :

atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m

atomic wieght, M = 95.96

Cell structure is BCC (Body Centred Cubic)

For BCC, we know that no. of atoms per unit cell, z = 2

and atomic radius, r =\frac{a\sqrt{3} }{4}

so, a = \frac{4r}{\sqrt{3}}

m = mass of each atom in a unit cell

mass of an atom = \frac{M}{N_{A} },

where, N_{A} is Avagadro Number = 6.02×10^{23}

volume of unit cell = a^{3}

density, ρ = \frac{mass of unit cell}{volume of unit cell}

density, ρ = \frac{z\times M}{a^{3}\times N_{A}}

ρ = \frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}

ρ = 10.215gm/cm^{3}

5 0
3 years ago
A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-sta
KIM [24]

Answer:

Temperature distribution is T(x)=\dfrac{q}{k}(L-x)+T

Heat flux=q

Heat rate=q A  

Explanation:

We know that for no heat generation and at steady state

\dfrac{\partial^2 T}{\partial x^2}=0

\dfrac{\partial T}{\partial x}=a

T(x)=ax+b

a and are the constant.

Given that heat flux=q

We know that heat flux given as

q=-K\dfrac{dT}{dx}

From above we can say that

a= -\dfrac{q}{K}

Alos given that when x= L temperature is T(L)=T

T= -\dfrac{q}{K}L+b

b=T+\dfrac{q}{K}L

So temperature T(x)

T(x)=-\dfrac{q}{K}x+T+\dfrac{q}{K}L

T(x)=\dfrac{q}{k}(L-x)+T

So temperature distribution is T(x)=\dfrac{q}{k}(L-x)+T

Heat flux=q

Heat rate=q A         (where A is the cross sectional area of wall)

   

6 0
3 years ago
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