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dezoksy [38]
2 years ago
9

vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Repo

rt your answer with proper units and 3 sig figs.
Engineering
1 answer:
DanielleElmas [232]2 years ago
5 0

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= density\times g\times \frac{h}{2}

By putting values, we get

= 1000\times 9.8\times \frac{12.2}{2}

= 1000\times 9.8\times 6.1

= 59780

hence,

The average force will be:

= Pressure\times Area

= 59780\times 3.6\times 12.2

= 2625537 \ N

Or,

= 2625 \ kN

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Please help me with this question​
kap26 [50]

Answer:

The current through each lamp is 0.273 Amperes

Power dissipated in each lamp is 0.082W

Explanation:

Battery v = 1.5 V

Each lamp has resistance, r = 1.1 Ohms

The 5 lamps in series will therefore have total resistance, R = 5 * 1.1 = 5.5 Ohms

The current through each lamp, I = v/R = 1.5 / 5.5 = 0.273 Amperes

Power dissipated in each lamp = I² * r = 0.273² * 1.1 = 0.082W

3 0
3 years ago
A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

4 0
4 years ago
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
How are speed and acceleration related
erica [24]
R = distance

dr/dt speed or with a direction, velocity

d(dr/dt)/dt = the time derivative of the velocity is called acceleration.

Speed is a scalar. Acceleration is a vector.
6 0
3 years ago
Read 2 more answers
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
natima [27]

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0
3 years ago
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