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nata0808 [166]
3 years ago
12

A rectangular channel 2 m wide carries 3 m3 /s of water at a depth of 1.2 m. If an obstruction 40 cm wide is placed in the middl

e of this channel, find the elevation of the water surface at the constriction (Note that Q remains unchanged but q changes). What is the minimum width of the constriction that will not cause a rise in the water surface upstream?
Engineering
1 answer:
Marta_Voda [28]3 years ago
8 0

harden you could either me or leave

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Xharden you could either me or leave

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BBB

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GO WATCH AFTER OUT NOW RADDED RB

harden you could either me or leave

harden you could either me or leave

harden you could either me or leave

harden you could either me or leave

GO WATCH AFTER OUT NOW RADDED R

harden you could either me or leave

GO WATCH AFTER OUT NOW RADDED R

harden you could either me or leave

harden you could either me or leave

GO WATCH AFTER OUT NOW RADDED R

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A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =
Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

F=ma for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that  W = mg

For simplicity we work with g =9.807 \frac{m}{s^{2}} despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula g = a-bz the "normal" or "standard" weight of an object is given by W = mg = ma when z = 0, so we need to find the value of z that makes W = m(a-bz) = 0.997ma meaning that the original weight decrease by a 0.30%, so now we operate...

m(a-bz) = 0.997ma now we group like terms on the same sides ma(1-0.997) = mbz we cancel equal tems on both sides and obtain that z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m

7 0
3 years ago
Water enters a tank from two pipes, one with a flow rate of 0.3 kg/s and the other with a flow rate of 0.1 kg / s. The tank has
stiks02 [169]

Answer:

total amount of water after 2 min will be 84.4 kg/s

Explanation:

Given data:

one tank inflow = 0.1 kg/s

2nd tank inflow = 0.3 kg/s

3rd tank outflow = 0.03 kg/s

Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s

From third point, outflow is 0.03 kg/s

Therefore, resultant in- flow = 0.4 - 0.03

Resultant inflow is  = 0.37 kg/s

Tank has initially 40 kg water

In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg

So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg

8 0
3 years ago
A __________ is a single lane used by drivers to enter and exit a freeway. a) Climbing lane b) Weave lane c) Thru lane d) Offset
Rufina [12.5K]

A Weave lane is a single lane used by drivers to enter and exit a freeway.

<h3>What is this lane about?</h3>

Weave lanes are known to be lanes that acts as an entrance and exits for a lot of cars in highways.

Hence, one can say that A Weave lane is a single lane used by drivers to enter and exit a freeway.

Learn more about Weave lane from

brainly.com/question/10828527

#SPJ1

6 0
1 year ago
public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fracti
True [87]

Answer:

The Full details of the answer is attached.

7 0
3 years ago
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