A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity

Answers

Answer 1

Answer:

Answer:

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

Explanation:

From Open Channel Theory we know that critical depth of the rectangular channel ( $y_(c)$ ), measured in meters, is calculated by using this equation:

$y_(c) = \sqrt[3]{(\dot V^(2))/(g\cdot b^(2)) }$ (Eq. 1)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$b$ - Channel width, measured in meters.

If we know that $\dot V = 15\,(m^(3))/(s)$ , $g = 9.807\,(m)/(s^(2))$ and $b = 2\,m$ , then the critical depth is:

$y_(c) = \sqrt[3]{(\left(15\,(m^(3))/(s) \right)^(2))/(\left(9.807\,(m)/(s^(2)) \right)\cdot (2\,m)^(2)) }$

$y_(c) \approx 1.790\,m$

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity ( $v$ ), measured in meters, is obtained from this formula:

$v = (\dot V)/(b\cdot y_(c))$ (Eq. 2)

If we know that $\dot V = 15\,(m^(3))/(s)$ , $b = 2\,m$ and $y_(c) = 1.790\,m$ , then the flow velocity in the rectangular channel is:

$v = (15\,(m^(2))/(s) )/((2\,m)\cdot (1.790\,m))$

$v = 4.190\,(m)/(s)$

The flow velocity in the rectangular channel is 4.190 meters per second.

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

The percentage loss of the window is $E = 97.3$ %

Explanation:

From the question we are told that

The area of pane of glass is $A = 0.15 m^2$

The thickness is $d = 5mm = (5)/(1000) = 0.005m$

The thickness of the wall is $D = 0.15m$

The area of the wall is $a = 10m^2$

Generally the heat lost as a result of conduction of the window is

$Q_(window) = (j_(glass) * A * (\Delta T) )/(d)$

Where $j_(glass)$ is the thermal conductivity of glass which has a constant value of

$j_(glass) = 0.80 J/(s \cdot m \cdot C^o)$

Substituting values

$Q_(window) = ( 0.80 * 0.15 * (\Delta T) )/(0.005)$

$Q_(window) = 24 \Delta T$

Generally the heat lost as a result of conduction of the wall is

$Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)$

$j_(styrofoam)$ s the thermal conductivity of Styrofoam which has a constant value of $j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)$

Substituting values

$Q_(wall) = ( 0.010 * 10 * (\Delta T) )/(0.15)$

$Q_(wall) = 0.667 \ \Delta T$

Now the net loss of heat is

$Q_(net) = Q_(window) + Q_(wall)$

Substituting values

$Q_(net) = 24 + 0.667$

$Q_(net) = 24.667$

Now the percentage loss by the window is

$E = (Q_(window) )/(Q_(net)) * 100$

Substituting value

$E = (24)/(24 .667) * 100$

$E = 97.3$ %

B. Suppose R1 is a fuse which burns out due to a sudden surge of current, thus, it essentially becomes an open switch. How do the currents change after this?

Answers

Answer:

The currents becomes 0

Explanation:

when the fuse burns out due to a sudden surge of current and becomes an open switch (with a resistance of Infinity ∞) this automatically reduces the currents through it to zero

The Clausius inequality expresses which of the following laws? i. Law of Conservation of Mass ii. Law of Conservation of Energy iii. First Law of Thermodynamics iv. Second Law of Thermodynamics

Answers

Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow down by decelerating rate of 12 ft/sec2 . Will the truck be able to stop in time?

Answers

Answer:

The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Explanation:

The distance that the truck starts slowing down = 80 ft from the stop sign

Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.

u = initial velocity of the truck = 40 mph = 58.667 ft/s

v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)

x = horizontal distance covered during the deceleration

a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

0² = 58.667² + 2(-12)(x)

24x = 3441.816889

x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Corrected Question:

A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?

Answer:

The truck will not be able to stop in time.

Explanation:

==> First lets convert all variables to SI units

1 mph = 0.45m/s

40mph = 40 miles per hour = 40 x 0.45 m/s

40mph = 18m/s

1 ft = 0.3048m

80 ft = 80 x 0.3048m

80 ft = 24.38m

Also;

12ft/s² = 12 x 0.3048m/s²

12ft/s² = 3.66m/s²

==> Now, consider one of the equations of motion as follows;

v² = u² + 2as -----------------(i)

Where;

v = final velocity of motion

u = initial velocity of motion

a= acceleration/deceleration of motion

s = distance covered during motion

Using this equation, lets calculate the distance, s, covered during the acceleration;

We know that;

v = 0 [since the truck comes to a stop]

u = 40mph = 18m/s

a = -12ft/s² = -3.66m/s² [the negative sign shows that the truck decelerates]

Substitute these values into equation (i) as follows;

0² = 18² + 2 (-3.66)s

0 = 324 - 7.32s

7.32s = 324

s = $(324)/(7.32)$

s = 44.26m

The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m). This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.

Which of the following was an effect of world war 2 on agricultural industry

Answers

Answer:

Option C..Farmers saught new technology to help with the workload

hope this helped you

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Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

Answers

Answer:

Decrease to typical from utilizing lambda-decrease:

The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2

The of taking the terms is significant in lambda - math,

For the term, (λy, y×3)2, we can substitute the incentive to the capacity.

Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6

Presently the tem becomes, (λf λx f(f(fx)))6

The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.

Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.

In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.

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