Explain when it is appropriate to use Tier II?

Answer:

the force conveyed by the fibers is 947.93 lb-f

Explanation:

Given the data in the question;

V_f = 80% = 0.8

V_m = 1 - V_f = 1 - 0.8 = 0.2

Now,

length of fibre L_f = length of Nylon L_n

V_f = A_f × L_f = 0.8

V_m = A_n × L_n = 0.2

so

V_f/V_m = A_f/A_n = 0.8/0.2

A_f/A_n = 4

now, the strains in fibre is equal to strains in nylon

(P/AE)f = (P/AE)n

P_f/A_fE_f = P_n/A_nE_n

P_f = (A_f/A_n)(E_f/E_n)(P_n)    

P_f = ( 4 )( 131 / 2.8 )(Pn)  

P_f = 187.14Pn

and P_n = Pf / 187.14

Hence

given that P_total = 953 lb-f

P_f + P_n = 953

P_f + ( P_f / 187.14 ) = 953

P_f( 1 + ( 1 / 187.14 ) ) = 953

P_f( 1.00534359 = 953

P_f = 953 / 1.00534359

P_f = 947.93 lb-f

Therefore, the force conveyed by the fibers is 947.93 lb-f

Answer: look at the screenshot

The power and thermal efficiency of this engine is equal to 241 Kilowatts and 48.2% respectively.

First of all, we would determine the thermal efficiency of this engine by applying the following formula:

[tex]\eta= 1-\frac{1}{r_p^{k-1/k}} \\\\\eta= 1-\frac{1}{10^{1.4-1/1.4}} \\\\\eta= 1-\frac{1}{10^{0.4/1.4}}\\\\\eta= 1-\frac{1}{10^{0.4/1.4}}\\\\\eta= 1-\frac{1}{10^{0.2857}}\\\\\eta = 0.482[/tex]

Thermal efficiency = 48.2%.

Now, we can determine net power output as follows:

[tex]W_{out}=nq_{in}\\\\W_{out}= 0.482 \times 500\\\\W_{out}=241\;kW.[/tex]

Power = 241 Kilowatts.

Read more on thermal efficiency here: https://brainly.com/question/13577244

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Answer:

The thrust of the engine calculated using the cold air is 34227.35 N

Explanation:

For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as

[tex]\dot{m}=\rho AV_a[/tex]

Here

So the equation becomes

[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]

Now in order to find the flow from the fan, the Bypass ratio is used.

[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]

Here BPR is given as 8 so the equation becomes

[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]

Now the exit velocity is calculated using the total energy balance which is given as below:

[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]

Here

So the equation becomes

[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]

Rearranging the equation gives

[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]

Now using  the cold air approach, the thrust is given as follows

[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]

So the thrust of the engine calculated using the cold air is 34227.35 N

Answer:

The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )

Explanation:

width of channel = 3.0 m

Flow rate = 5 m^3/s

Normal depth = 0.50 m

Flow encounters a dam rise of 0.25 m

To know if the hydraulic jump will occur

we will Determine the new normal depth

Y2 = 3.77m

Yc ( critical depth )= 0.66m

Attached below is the detailed solution

Answer:

"7.654 mm" is the correct solution.

Explanation:

According to the question,

Now,

As we know,

The Elongation,

⇒ [tex]E=\frac{\sigma}{e}[/tex]

       [tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]

or,

⇒ [tex]\delta=\frac{PL}{AE}[/tex]

By substituting the values, we get

 [tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]

then,

⇒ [tex]D^2=58.587[/tex]

     [tex]D=\sqrt{58.587}[/tex]

         [tex]=7.654 \ mm[/tex]

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y  ----- ( 1 )

M ( bending moment ) = R * length of span / 2

                                     = (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

                                    = ( 2 )*( d )^3  / 12 =  2d^3 / 12

b = 2 in ,  d = ?

length of span = 4 * 12 = 48 inches

R = P  / 2 =  240 * 10^3 / 2 =   120 * 10^3 Ib

y ( centroid distance ) = d / 2  inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

                = ( 288 * 10^4 * 12 ) / 2d^3 )  * d / 2

940300  = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d  ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188   ∴ Value of d ≈ 3.03 in

Answer:

so hard it is

Explanation:

I don't know about this

please mark as brainleast

byýyy

Answer:

a. 89.5 N b. 1.59 N

Explanation:

a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?

Using Boyle's law, we find the final pressure at the compressed state given that the initial pressure is atmospheric pressure

So, P₁V₁ = P₂V₂ where P₁ = initial atmospheric pressure in syringe = 1 atm = 1.013 × 10⁵ N/m²,V₁ = initial volume of syringe = 5 ml, P₂ = final pressure in syringe at compression and V₂ = final volume of syringe = 1 ml

So, making P₂ subject of the formula, we have

P₂ = P₁V₁/V₂

Substituting the values of the variables into the equation, we have

P₂ = P₁V₁/V₂

P₂ = 1.013 × 10⁵ N/m² × 5 ml/1 ml

P₂ = 1.013 × 10⁵ N/m² × 5

P₂ = 5.065 × 10⁵ N/m²

Since pressure, P = F/A where F = force and A = cross-sectional area of syringe = πd²/4 where d = diameter of syringe = 1.5 cm = 1.5 × 10⁻² m.

So, F = PA

F = P₂πd²/4

substituting the values of the variables into the equation, we have

F = P₂πd²/4

F = 5.065 × 10⁵ N/m²π(1.5 × 10⁻² m)²/4

F = 5.065 × 10⁵ N/m²π(2.25 × 10⁻⁴ m²)/4

F = 35.8 × 10/4 N

F = 8.95 × 10

F = 89.5 N

b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?

Since the pressure in the syringe after compression is constant, we have

P₂ = F₁/A₁ where F₁ = force exerted on thumb and A₁ = cross-sectional area of  opening of syringe = πd₁²/4 where d = diameter of opening of syringe = 2 mm = 2 × 10⁻³ m.

So, F₁ = P₂A₁

F = P₂πd₁²/4

substituting the values of the variables into the equation, we have

F = P₂πd²/4

F = 5.065 × 10⁵ N/m²π(2 × 10⁻³ m)²/4

F = 5.065 × 10⁵ N/m²π(4 × 10⁻⁶ m²)/4

F = 15.91 × 10⁻¹

F = 1.591 N

F ≅ 1.59 N

Answer:

sorry i don't know

Explanation:

Answer:

Explanation:

You'll still need wrenches, sockets and screwdrivers, but there are other things that it may be necessary to buy to complete the work.

Spring Compressor. One part of suspension repair is replacing coil springs.  

Hydraulic Puller. -Hydraulic pullers are used to remove shaft-fitted parts (bearings or couplings). Pullers use a controlled hydraulic force in an effective way and can quickly separate (especially compared to the manual alternative) the parts.

CV Boot Tool. he CV Boot is a ribbed, rubber flexible boot that keeps water and dirt out of the joint and the special grease inside the joint.

Torque Wrench. .A torque wrench is a tool used to apply a specific torque to a fastener such as a nut, bolt, or lag screw. ... A torque wrench is used where the tightness of screws and bolts is crucial.

Ball Joint Separator. This tool is used to separate the ball joint from the spindle support arm. It works on many domestic and import front wheel drive vehicles and is adjustable up to 2" for different size ball joints.

Strut Nuts.  

Tie Rod Puller.

etc....

Answer:

0.1365 m^3

Explanation:

thickness of upper aquifer = 500 ft

lower aquifer head drop = 150 ft

area of ground water basin = 12 m^2

specific yield of upper unit = 0.12

Storativity of lower unit = 4 * 10^-4

determine the amount of recoverable ground water

first step : calculate volume of unconfined aquifer  

                = 12 * 500/5280   = 1.1364 miles^3

The recoverable volume of water from unconfined aquifer

= 1.1364 * 0.12  = 0.1364 miles^3

next : calculate volume of confined aquifer

       = 12 * 150/5250 = 0.341 miles^3

The recoverable volume of water from confined aquifer

     = 0.341 * ( 4 * 10^-4 )  = 1.364 * 10^-4 miles^3

Hence the amount of recoverable ground water in the basin

= ∑ recoverable ground water from both aquifer

 = 0.1365 m^3

Answer:

O The mix is uniform in color