A tank has a gauge pressure of 552 psi. The cover of an inspection port on the tank has a surface area of 18 square inches. What is the total force the cover is experiencing.

Answers

Answer 1

Answer:

44197.55 N

Explanation:

From the question,

Pressure of the pressure guage (P) = Total force experienced by the cover (F)/Area of the cover (A)

P = F/A................ Equation 1

make F the subeject of the equation

F = P×A............... Equation 2

Given: P = 552 psi = (552×6894.76) = 3805907.52 N/m², A = 18 square inches = (18×0.00064516) = 0.01161288 m²

Substitute these values into equation 2

F = ( 3805907.52×0.01161288)

F = 44197.55 N


Related Questions


A source current of 10 mA is supplied to a parallel circuit consisting of the following resistors three resistors, a 2200 a 500 and a 1KO. What is the source voltage required to
supply the current

Answers

It’s is 2000 because it is 20000


3.Which of the following drawings are matched with the project specifications to form the bulk of the contract document?

Answers

Can you show us the drawings

A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. During the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.

Answers

Answer:

LOS = A

Explanation:

Given all the parameters the level of service as seen from the attached graph

is LOS =  A

To determine the LOS from the attached graph

calculate the trial value of Vp

Vp = V / PHF

     = (100 + 150) / 0.95  =  263 pc/h

since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1

next we will calculate the flow rate

flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]

             Fhr  = 1 / 1.035 = 0.966 ≈ 1

next calculate the real value of Vp

Vp = V / ( PHF * N * Fhr * Fp )

     = ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )

Vp ≈ 126 pc/h/In

Next calculate the density

D = Vp /  S  =  126 / ( 45 * 1.61 )  = 1.74 pc/km/In

Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=500 (10^6)mm^4

Answers

Answer:

hello your question is incomplete attached below is the complete question

answer :

Slopes : B = 180 mm , C = 373 mm

Deflection: B = 0.0514 rad ,  C = 0.077 rad

Explanation:

Given data :

I = 500(10^6) mm^4

E = 70 GPa

The M / EI  diagram is attached below

Deflection angle at B

∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI

                 = 1800 / ( 500 * 70 ) = 0.0514 rad

slope at B

ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI

                 = 6300 / ( 500 * 70 ) = 0.18 m = 180 mm

Deflection angle at C

∅C = ∅CA = [ 1800 + 300*3 ] / EI

                 = 2700 / ( 500 * 70 )

                 = 2700 / 35000 = 0.077 rad

Slope at C

ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]

     = 13050 / 35000 = 373 mm

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Answers

Answer:

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Methane (CH4) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O2 and 79% N2 on a molar basis.

Required:
Determine the dew point temperature of the products, in K.

Answers

ATM enters a furnace operating at steady state and burns completely

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 fracture toughness of 92 MPa m (klein.)) and a yield strength of 900 MPa (65270 psi). The flaw size resolution limit of the flaw detection apparatus is 3 mm (0.1181 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.15, what is the critical flaw length

Answers

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness [tex]K_{tc[/tex] = 92 Mpa√m

yield strength σ[tex]_y[/tex] = 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 [tex]a_c[/tex] = 1/π( [tex]K_{tc[/tex] / Yσ )²

we substitute

[tex]a_c[/tex] = 1/π( 92 Mpa√m / (1.15 × 450 Mpa  )²

[tex]a_c[/tex] = 1/π( 92 Mpa√m / (517.5 Mpa  )²

[tex]a_c[/tex] = 1/π( 0.177777  )²

[tex]a_c[/tex] = 1/π( 0.03160466 )

[tex]a_c[/tex] = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ [tex]a_c[/tex] = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 6 when 16.0% of the crack length is removed.

Answers

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

([tex]p_t[/tex])₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2

where ([tex]p_t[/tex])₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]  

divide both sides by 2(σ₀)₁

[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex]  =  36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]

1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂

([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁

([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

An acid-base indicator is usually a weak acid with a characteristic color in the protonated (acid) and deprotonated (conjugate base) forms. In this assignment, you will monitor the color of an acetic acid solution containing Bromocresol Green as an indicator, as the pH is changed and then you will use your data to calculate the ionization constant, Ka, for the bromocresol green indicator and compare to an accepted value.
1. Start Virtual Chemlab, select Acid-Base Chemistry, and then select lonization Constants of Weak Acids from the list of assignments. The lab will open in the Titration laboratory. Bottles of 0.1104 M NaOH and 0.1031 M HAC (acetic acid) will be on the lab bench. The buret will be filled with the NaOH solution and a beaker containing 10.00 mL of the HAc solution will be on the stir plate. The stir plate will be on, Bromocresol Green indicator will have been added to the beaker, and a calibrated pH probe will also be in the beaker so the pH of the solution can be monitored.
2. What is the color and pH of the solution?
3. On the buret, the horizontal position of the orange handle is off for the stopcock. Open the stopcock by pulling down on the orange handle. The vertical position delivers solution the fastest with three intermediate rates in between (slow dropwise, fast dropwise, and slow stream). Turn the stopcock to the second position or fast drop-wise addition. Observe the color of the solution and close the stopcock when you observe the change in color by double clicking on the center of the stopcock.
4. What is the color and pH of the solution now?
5. Continue to add NaOH as before or at a faster rate. What is the final color and pH of the solution after all of the NaOH is added?
6. An acid-base indicator is usually a weak acid with a characteristic color in the protonated and deprotonated forms. Because bromocresol green is an acid, it is convenient to represent its rather complex formula as HBOG. HBOG ionizes in water according to the following equation:
HBOG + H2O = BCG + H3O+
(yellow) (blue)
The K. (the equilibrium constant for the acid) expression is:
Ka = [BCG-][H3O+]/[HBCG)
When [BCG-[ = [HBCG), then Ka = [H3O+). If you know the pH of the solution, then the [H3O+] and Ka can be determined.
What would be the color of the solution if there were equal concentrations of HBCG and BCG-?
What is the pH at the first appearance of this color?
What is an estimate for the Ka for bromocresol green?

Answers

An acid base indicator is usually a weak acid with Terroristic color

14. The top plate of the bearing partition
I
a. laps the plate of the exterior wall.
b. is a single member.
c. butts the top plate of the exterior wall.
d. is applied after the ceiling joists are
installed.

Answers

Answer:

d. is applied after the ceiling joists are

installed.

An air conditioning system is to be filled from a rigid container that initially contains 5 kg of saturated liquid at 24° Celsius the valve connecting this container to the air conditioning system is not open until the mass in this container is .25 Cal and the quality is going 506 at which time the valve is closed during this time only saturated liquid R134a flows from the container presuming that the process is isothermal wild the valve is open.

Required:
Determine the final quality of the R-134a in the container and the total heat transfer.

Answers

And air-conditioning system is to be filled for my ridge the containerBut that internally contains 5 kgDetermine the final quality of the arm 134

A piston–cylinder device contains 0.8 kg of steam at 300°C and at pressure of 800 kPa. Steam is cooled at constant pressure until one-half of the mass condenses. The final temperature is

a.
178°C

b.
184°C

c.
195°C

d.
None of these

e.
167°C

f.
170°C

Answers

Answer:

d

Explanation:

d.

None of these

Identify the following formulas:
1. Slope Formula
2. Slope-Intercept Form
3. Standard Form
4. Point-Slope Formula

Answers

1. In the image below
2. y=mx+b
3. Ax+By=C (I think)
4. y-y1=m(x-x1)

Create a 6-bit full subtractor that uses the Borrow method to subtract two 6-bit binary numbers. You can use the proper basic sub-circuit.

Answers

Create a six pack for subtract or that uses to borrowMessage you subtract 26 fit

An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% level of signigicance, is that th mean shear strenfth of spot welds is at least 3.1 MPa. The engineer randomly selcts 15 Welds and measures the shear strength is 3.07 MPa with a sample standard deciation of 0.069 MPa. Which of the following statment is true?
a) The null hypothesis should not be rejected.
b) The null hypothesis should be rejected.
c) The alternate hypothesis should be rejected.
d) The null and alternate hypotheses are equally likely.

Answers

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources

Answers

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

We are given a CSP with only binary constraints. Assume we run backtracking search with arc consistency as follows. Initially, when presented with the CSP, one round of arc consistency is enforced. This first round of arc consistency will typically result in variables having pruned domains. Then we start a backtracking search using the pruned domains. In this backtracking search we use filtering through enforcing arc consistency after every assignment in the search.
Which of the following are true about this algorithm?
a) If after a run of arc consistency during the backtracking searchwe end up with the filtered domains of allof the not yetassigned variables being empty, this means the CSP has nosolution.
b) If after a run of arc consistency during the backtracking searchwe end up with the filtered domain of oneof the not yetassigned variables being empty, this means the CSP has nosolution.
c) None of the above.

Answers

We are given a CSP with only binary can concentrate assume we run backtrackingSearch with ARC

An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through the collector at a flow rate of 0.06 l/s. The ambient temperature is 8 C and the exit temperature is 49 C. Determine the overall heat loss coefficient.

Answers

Answer:

- 14.943 W/m^2K  ( negative sign indicates cooling )

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

Calculate the overall heat loss coefficient

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( [tex]\frac{49-8}{60-8}[/tex] ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

Can someone put each letter by the correct word for my automotive class !

Answers

Answer:

L = spindle

M = lower ball joint

part without the letter showing = steering knuckle

Explanation:

calculate force and moment reactions at bolted base O of overhead traffic signal assembly. each traffic signal has a mass 36kg, while the masses of member OC and AC are 50Kg and 55kg, respectively. The mass center of mmber AC at G.​

Answers

Answer:

The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F

Explanation:

I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.

what is the most common type of suspensions system used on body over frame vehicles?

Answers

Answer:

Engine

Explanation:

Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.

What is a Semi-independent suspension?

Semi-independent suspension give the front wheels some individual movement.

This suspension only used in rear wheels.

Thus, the correct option is Semi-independent suspension

Learn more about Semi-independent suspension

https://brainly.com/question/23838001

#SPJ2

.If aligned and continuous carbon fibers with a diameter of 6.90 micron are embedded within an epoxy, such that the bond strength across the fiber-epoxy interface is 17 MPa, and the shear yield strength of the epoxy is 68 MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is 3960 MPa.

Answers

Answer:

the required minimum fiber length is 0.80365 mm

Explanation:

Given the data in the question;

Diameter D = 6.90 microns = 6.90 × 10⁻⁶ m

Bond strength ζ = 17 MPa

Shear yield strength ζ[tex]_y[/tex] = 68 Mpa

tensile strength of carbon fibers [tex]6t_{fiber[/tex] = 3960 MPa.

To determine the minimum fiber length we make use of the following relation;

L = ([tex]6t_{fiber[/tex] × D) / 2ζ

we substitute our given values into the equation;

L = ( 3960 × 6.90 × 10⁻⁶) / (2 × 17 )

L = 0.027324 / 34

L = 0.000803647 m

L = 0.000803647 × (1000) mm

L = 0.80365 mm

Therefore, the required minimum fiber length is 0.80365 mm

5. Which of these materials in a shop contain metals and toxins and can pollute the environment? A) Antifreeze B) Solvents C) Batteries D) All of the above

I say it's D) All of the above​

Answers

As a multiple choice the only correct answer is D

Answer:

D

Explanation:

All of the above

Explain two reason why it is important for DG08 Engineering to refer to electronic component pin configuration specifications when designing and building printed circuit boards?

Answers

Answer:

refer to aja

Explanation:

Explain biometric senser.​

Answers

Biometric sensors are used to collect measurable biological characteristics from a human being, which can then be used in conjunction with biometric recognition algorithms to perform automated person identification.

Answer:

Biometric sensors are used to collect measurable biological characteristics (biometric signals) from a human being, which can then be used in conjunction with biometric recognition algorithms to perform automated person identification.

Al ejercer una fuerza de 50N sobre un resorte elastico esto se alarga desde los 15 cm hasta los 60cm¿cual es la constante elastica del resorte?

Answers

Answer:

Constante de resorte = 1.1 N/m

Explanation:

Dados los siguientes datos;

Fuerza = 50N

Extensión = 60cm - 15cm = 45cm

Para encontrar la constante del resorte;

Matemáticamente, la fuerza ejercida para estirar un resorte viene dada por la fórmula;

Fuerza = constante de resorte * extensión

Sustituyendo en la fórmula, tenemos;

50 = constante de resorte * 45

Constante de resorte = 50/45

Constante de resorte = 1.1 N/m

Please help me answer this engineering question

Answers

https://www.bartleby.com/solution-answer/chapter-26-problem-5sq-electric-motor-control-10th-edition/9781133702818/what-is-meant-by-the-low-or-poor-starting-economy-of-a-primary-resistor-starter/fb257667-8e6f-11e9-8385-02ee952b546e this website has 26 solutions maybe this will help

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T

Answers

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 +  (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C

You will be hiking to a lake with some of your friends by following the trails indicated on a map at the trailhead. The map says that you will travel 1.7 mi directly north, then 2.7 mi in a direction 36° east of north, then finally 1.7 mi in a direction 15° north of east. At the end of this hike, how far will you be from where you started, and what direction will you be from your starting point?

Answers

Answer: 64.58



Explanation:

banana with an average mass of 0.15 kg and average specific heat of 3.35 kJ/kg · °C is cooled from 20°C to 5°C. The amount of heat transferred from the banana is

a.
62.1 kJ

b.
7.5 kJ

c.
None of these

d.
6.5 kJ

e.
0.85 kJ

f.
17.7 kJ

Answers

Answer:

The amount of heat transferred from the banana is (-)7.54 KJ

Explanation:

As we know,

[tex]Q = m*c*\Delta T[/tex]

Q = Amount of heat transferred

m = mass of banana

[tex]T_2 = 5[/tex] degree Celsius

[tex]T_1 = 20[/tex] degree Celsius

The amount of heat transferred from the banana =

[tex]0.15 * 3.35 * (5 -20)\\-7.54[/tex]KJ (negative sign represents reduction in heat energy)

Other Questions
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