A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that rises 0.25 m above the channel bottom. Will a hydraulic jump occur?

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:

sorry i don't know

Explanation:

Answer:

so hard it is

Explanation:

I don't know about this

please mark as brainleast

byýyy

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:

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:

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