This is it dont anwser this is for my other account

A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 83,051 N , and experien

umka2103 [35]

Answer:

the answer is attached with required units.

Explanation:

5 0

2 years ago

Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

cricket20 [7]

Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

6 0

3 years ago

A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b

Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= $\frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}$

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= $\frac{8640*100-8640*20}{20*10^5}$

k= 0.35 per day

6 0

2 years ago

Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,

Tems11 [23]

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

F

=

(

20

×

10

−

9

C

)

(

10

m/s

)

(

5

×

10

−

5

T

)

=

1

×

10

−

11

(

C

⋅

m/s

)

(

N

C

⋅

m/s

)

=

1

×

10

−

11

N

6 0

2 years ago

Read 2 more answers

If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical

vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 × $10^{9}$ N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$ .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × $10^{-3}$ m

so now put value in equation 1 we get

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$

( σc ) = $\sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}$

( σc ) = 27.396615 × $10^{6}$ N/m²

so critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

6 0

2 years ago