All categories

3 years ago

Which branch of engineering studies the physical behavior of metallic elements?

Engineering

2 answers:

olga55 [171]3 years ago

6 0

Material engineering studies the physical behavior of metallic elements.

Answer: Option C

<u>Explanation: </u>

Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.

Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.

steposvetlana [31]3 years ago

3 0

Answer:

your answer should be c.

You might be interested in

A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete

Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt) where now t=t₂

$d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min$

4 0

3 years ago

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this

tresset_1 [31]

Answer:

a)COP=5.01

b) $W_{in}=2.998$ KW

c)COP=6.01

d) $Q_R=17.99 KW$

Explanation:

Given

$T_L$ = -12°C, $T_H$ =40°C

For refrigeration

We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

$COP=\dfrac{T_L}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{261}{313-261}$

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that $COP=\dfrac{RE}{W_{in}}$

RE is the refrigeration effect

5.01= $\dfrac{15}{W_{in}}$

b) $W_{in}=2.998$ KW

For heat pump

So COP of heat pump is given as follows

$COP=\dfrac{T_h}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{313}{313-261}$

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at low temperature+work in put

$Q_R=Q_A+W_{in}$

Given that $Q_A=15$ KW

We know that $COP=\dfrac{Q_R}{W_{in}}$

$COP=\dfrac{Q_R}{Q_R-Q_A}$

$6.01=\dfrac{Q_R}{Q_R-15}$

d) $Q_R=17.99 KW$

5 0

3 years ago

three balls each have a mass m if a has a speed v just before a direct collision with B determine the speed of C after collision

ratelena [41]

Answer:

Vc2= V(l+e) ^2/4

Vg2= V(l-e^2)/4

Explanation:

Conservation momentum, when ball A strikes Ball B

Where,

M= Mass

V= Velocity

Ma(VA)1+ Mg(Vg)2= Ma(Va)2+ Ma(Vg)2

MV + 0= MVg2

Coefficient of restitution =

e= (Vg)2- (Va)2/(Va)1- (Vg)1

e= (Vg)2- (Va)2/ V-0

Solving equation 1 and 2 yield

(Va)2= V(l-e) /2

(Vg)2= V(l+e)/2

Conservative momentum when ball b strikes c

Mg(Vg)2+Mc(Vc)1 = Mg(Vg)3+Mc(Vc)2

=> M[V(l+e) /2] + 0 = M(Vg)3 + M(Vc) 2

Coefficient of Restitution,

e= (Vc)2 - (Vg)2/(Vg)2- (Vc)1

=> e= (Vc)2 - (Vg)2/V(l+e) /2

Solving equation 3 and 4,

Vc2= V(l+e) ^2/4

Vg2= V(l-e^2)/4

8 0

3 years ago

Read 2 more answers

Question 1 of 8.

Rasek [7]

The answer seems pretty obvious, all of the above

8 0

3 years ago

Read 2 more answers

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago