Fixed rate mortgage offer:

A 3-m-high, 11-m-wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B. Determine the hyd

serg [7]

Answer:

its a

Explanation:

its a

5 0

3 years ago

A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st

Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

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5 0

1 year ago

A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna

Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

7 0

3 years ago

Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Repor

REY [17]

Answer:

$T_2=315.69k$

Explanation:

Initial Temperature $T_1=500K$

Initial Pressure $P_1=1000kPa$

Final Pressure $P_2=200kPa$

Generally the gas equation is mathematically given by

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}$

Where

n for $CO=1.4$

Therefore

$\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}$

$T_2=315.69k$

7 0

3 years ago

Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain f

BabaBlast [244]

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness $K_{tc$ = 92 Mpa√m

yield strength σ $_y$ = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length $a_c$ = 1/π( $K_{tc$ / Yσ )²

we substitute

$a_c$ = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²

$a_c$ = 1/π( 92 Mpa√m / (517.5 Mpa )²

$a_c$ = 1/π( 0.177777 )²

$a_c$ = 1/π( 0.03160466 )

$a_c$ = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ $a_c$ = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

5 0

3 years ago