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olya-2409 [2.1K]
3 years ago
11

. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

de slope of 1:2 (v:h) with a discharge of 10 (m3/s). The slope is 0.0004 and Manning's roughness is 0.015.
Engineering
1 answer:
Over [174]3 years ago
5 0

Answer:

y  ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

v =1/n R^{2/3} s^{1/2}

where R is hydraulic radius

S = bed slope

Q = Av =A 1/n R^{2/3} s^{1/2}

A = 1/2 \times (B+B+4y) \times y =(B+2y) y

R =\frac{A}{P}

P is perimeter =  (B+2\sqrt{5} y)

R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}

Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}

solving for y100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}

solving for y value by using iteration method ,we get

y  ≈ 2.5

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In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise
lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

\dfrac{N_{sun}-N}{N_{ring}-N}=-R

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

\dfrac{N_{sun}-N}{N_{ring}-N}=-R

\dfrac{5-N}{12-N}=-8

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0
3 years ago
From the hazards blisted below, determine which could require foot protection​
MrRa [10]
Foot protection could be required with Sharps, Slippery areas, Hazardous liquids,and Falling objects.

Hope this helps!!
8 0
3 years ago
A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of th
fenix001 [56]

Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

So

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

6 0
3 years ago
Mary wants to study the codes that provide regulations to safeguard against fires and explosions arising from the storage and ha
MrRa [10]
I believe the answer you’re looking for is A. International Fire Code
6 0
3 years ago
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A roller support acts like a contact boundary condition as it can produce a reaction force as a push response to a body but will
Zanzabum

Answer:

a) True

Explanation:

Roller can provide reaction for push support but it can not provide reaction for pull support.  

From the free body diagram of roller and hinge support we can easily find that ,Roller providing vertical reaction and can not provide horizontal reaction.

On the other hand hinge support can provide reaction in both the direction.

So we can say that roller can not proved reaction for pull support.

6 0
3 years ago
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