All categories

2 years ago

Join my talkroom it is called CJ

Engineering

2 answers:

sashaice [31]2 years ago

7 0

Answer: heyy

Explanation:

k0ka [10]2 years ago

5 0

Answer:

Hello

Explanation:

You might be interested in

Training is a way for employers to provide ____ to enable employees to protect themselves and others from injuries.

Andreyy89

Answer:

Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.

Explanation:

5 0

2 years ago

Read 2 more answers

Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate

Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy + flow work

$E = \dot m ( \frac{v^2}{2] + zg + p\nu)$

$E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})$

$57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})$

solving for flow rate

$\dot m = 99.977we know that [tex]\dot m = \rho AV$

$\dot m = 780 \frac{\pi}{4} D^2\times 16$

solving for d

$99.97 = 780 \times \frac{\pi}{4} D^2\times 16$

d = 0.090 m

so radius = 0.045 m

3 0

3 years ago

A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg

Eva8 [605]

Answer:

$\dot{w}$ = -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

$\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}$

Given that tank is insulated so $\dot{Q}=0$ and no mass is leaving so

$\dot{m_e}=0$

$\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt$

$m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t$

Mass conservation $m_2-m_1=m_e-m_i$

$m_1,m_2$ is the initial and final mass in the system respectively.

Initially tank is evacuated so $m_1=0$

We know that for air $u=C_vT ,h=C_p T$ , $P_2v_2=m_2RT_2$

$m_2=0.42 kg$

So now putting values

$0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300$

$\dot{w}$ = -0.303 KW

3 0

3 years ago

What is the most common injury caused by working with machines unsafely?

VladimirAG [237]

Answer:

heyooo!!!!!!

Burns and scarring.

hope this helps!!!!

Explanation:

8 0

3 years ago

Read 2 more answers

Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str

alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

= 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

= 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

ρ_avg= mass of unit cell/volume of unit cell

ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

λ^3 = 4.37

6 0

3 years ago