the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate
This question is incomplete, the complete question is;
the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate ( in m³ ) if the rectangular channel is 20 m wide.
Answer:
the flow rate is 32.549 m³/sec
Explanation:
Given that
y₁ = 0.3 m
y₂ = 1.2 m
β = 20 m
Now for Rectangular Channel, we know that;
2q²/g = y₁y₂( y₁ + y₂)
where g = 9.81 m/s²
and q = Q/β
so
2(Q/β)²/g = y₁y₂( y₁ + y₂)
we substitute our given values
2(Q/20)²/9.81 = 0.3 × 1.2( 0.3 + 1.2)
2(Q²/400)/9.81 = 0.36(1.5)
2(Q²/400) = 0.54 × 9.81
Q²/400 = 5.2974 / 2
Q²/400 = 2.6587
Q² = 1059.48
Q = √1059.48
Q = 32.549 m³/sec
Therefore the flow rate is 32.549 m³/sec
Why is the reversed Carnot cycle executed within the saturation dome not a realistic model for refrigeration cycles?
Explanation:
The paths of warmth and job interactions are inverted by repeating the Carnot process. A Carnot fridge or a Carnot heating system is considered a fridge or heat engine that acts on the reverse Carnot cycle.
The Carnot process comprises liquid phase mixture compression and expanding, which limits mechanical components.
Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kept at a temperature of 80.6 °F year round using a natural gas heater. If you were to completely drain the pool and refill the pool with 50°F water, (blank) GJ (giga-Joules) of energy are required to to heat the water back to 80.6 °F. Note: The specific heat capacity of water is 4182 J/kg ⋅°C. The cost of natural gas per GJ is $2.844. It costs $ (blank) to heat the pool (to the nearest dollar).
Answer:
[tex]95.914\ \text{GJ}[/tex]
[tex]\$272.78[/tex]
Explanation:
m = Mass of water = 749511.5 kg
c = Specific heat of water = 4182 J/kg ⋅°C
[tex]\Delta T[/tex] = Change in temperature = [tex]80.6-50=30.6^{\circ}\text{F}[/tex]
Cost of 1 GJ of energy = $2.844
Heat required is given by
[tex]Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}[/tex]
Amount of heat required to heat the water is [tex]95.914\ \text{GJ}[/tex].
Cost of heating the water is
[tex]95.914\times 2.844=\$272.78[/tex]
Cost of heating the water to the required temperature is [tex]\$272.78[/tex].
who is the strongest avenger i say hulk but who knows at this point
Answer:
or is the strongest evenger she hulk
Explanation:
?????????
Answer:
Thor!
Explanation:
In Thor: Ragnarok he beat the Hulk in order for Hulk to win thor had to be electrocuted and in Avengers: Endgame Thor is seen holding open the "Floodgates" and withstanding the radiation from a dying star, also the fact that Thor is a god means that he is all powerful and the rightful heir to the throne to Asgard, plus the fact that he has defeated Loki multiple times a feat that not even the Hulk has done.
A brine solution is 26% salt with 70.0 kg of water evaporated per hour. To produce 195 kg of pure salt (0% moisture) per day, how long should the process operate each day and how much brine must be fed to the evaporator per hour?
The process should operate each day for ___ hours.
The amount of brine that must be fed to the evaporator is ___ kg/h.
Answer:
- the process should operate each day for 7.9286 hours
- the amount of brine that must be fed to the evaporator is 94.594 kg/hr
Explanation:
Given that;
concentration of brine = 26%
so water concentration will be 100% - 26% = 74%
for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)
so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr
Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr
Now for production of 195kg of pure dry salt, number of hours required will be
T = 195 / 24.5945 = 7.9286 hrs
Therefore the process should operate each day for 7.9286 hours.
Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg
Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr
Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr
A _________ is interesting only if the statistics computed from transactions covered by the rule are different than those computed from transactions not covered by the rule.
Answer: Quantitative association rule.
Explanation:
The quantitative rules of association apply to the basic type of rules of association which exists as X and Y, with X and Y consisting of a collection of numerical and/or categorical attributes. Unlike general association laws, where both the left and right sides of the law should have categorical (nominal or discrete) attributes, a numerical attribute must be included in at least one attribute of the quantitative association rule (left or right).
What is the primary reason that heating, ventilating, and air conditioning (HVAC) is critical to a data center
Answer:
It prevents hardware from overheating.
Explanation:
HVAC stands for Heating, ventilation and air conditioning. It may be defined as technology which provides both vehicular as well as indoor environmental comfort. It provides thermal comfort and also used in places where overheating of the equipment or object is not desired.
A data center is a dedicated space which is used to store computer systems and the associated components like telecommunication as well as storage system.
Now proper HVAC system is necessary in data center so as to ensure that the components or the computer system does not get over heated and gets damaged. Overheating of the storage system may lead to loss of valuable information and other important data.
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R14B, (6) G1B, (7) M8C3, (8) P1NSE, and (9) RL1.
(a) Determine the normal times in TMUs for these motion elements.
(b) What is the total time for this work element in sec
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
A blown fuse or tripped circuit breaker shows
Answer:
Your house is in need of a service upgrade, or it may indicate that your house has too few circuits.
Explanation:
It's a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits.
According to the Bureau of Labor Statistics, which occupation employed 4.2 million people – more than any other occupation – in 2010? Customer Service Representatives Retail Salespersons Cashiers Marketing Specialists
Answer:
Retail Salespersons
Explanation:
The Bureau of Labor Statistics or the BLS in short is the federal unit or agency of the United States Labor Department. It is the main stats finding agency of the United States federal government in the field of statistics and labor economics.
According to a publication published by the Labor Statistics Bureau, the retail salesperson sector of occupation employed the most people in the U.S. in the year 2010. It provided employment to about 4.2 million people.
Answer:
b. retail salespersons
Explanation:
Which metal has the ability to rust
A gold
B silver
C iron
D aluminum
Answer:
I got iron
Explanation:
on my plato test
Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says that a way to prevent galvanic corrosion is to reuse coated bolts. Who is right
Answer:
Technician A
Explanation:
Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.
5.5.1 Dual-edge detector A dual-edge detector is similar to a rising-edge detector except that the output is asserted for one clock cycle when the input changes from 0 to 1 (i.e., rising edge) and 1 to 0 (i.e falling edge). 1. Design a circuit based on the Moore machine and draw the state diagram and ASM chart. 2. Derive the HDL code based on the state diagram of the ASM chart. 3. Derive a testbench and use simulation to verify operation of the code.
The resistance of a 1,000-foot length of #6 AWG wire at a temperature of 25 degrees C is 0.4028 ohm. Thus, option B is correct.
What is prototype?It is a term used in a variety of contexts, including semantics, design, electronics, and software programming.A prototype is an early sample, model, or release of a product built to test a concept or process or to act as a thing to be replicated or learned from.
The statement that defines a prototype is a prototype is a working model of the proposed system. Say for instance a programmer creates a flowchart or an algorithm of an actual program. The flowchart or the algorithm is the prototype of the actual program. This in other words means that prototypes are models of a system. Hence, the statement that defines a prototype.
Therefore, The resistance of a 1,000-foot length of #6 AWG wire at a temperature of 25 degrees C is 0.4028 ohm. Thus, option B is correct.
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A commercial facility has a three-phase, 277 volt, 1200 Amp service. What is the required nominal transformer size needed for this building?
a. 750 kVA
b. 500 kVA
c. 575 kVA
d. 633 kVA
e. 1,000 kVA
Answer:
The answer is "Option e".
Explanation:
Given value:
[tex]\to v_{ph}=277 \ \ \ \ \ \ \ I_{ph} =1200\\\\[/tex]
[tex]vA= v_{Ph} I_{ph} \ \ \ \to for \ single \ phase\\\\=3v_{ph} I_{ph} \ \ \ \ \ \to for \ 3 - \phi \\\\s = 3\times 277 \times 1200 \\\\[/tex]
[tex]=997.200 \ K VA\\\\= 1000 \ k VA[/tex]
sin x +√3 coax= √2
Answer:
The two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12
Explanation:
The given equation is
Sin x +√3 Cosx= √2
Upon dividing the equation by 2 we get
[tex]\frac{1}{2}Sinx + \frac{\sqrt{3} }{2}Cosx = \frac{\sqrt{2} }{2}[/tex]
Sin([tex]\frac{pi}{6}[/tex])*Sinx + Cos([tex]\frac{pi}{6}[/tex])*Cosx = [tex]\frac{1}{\sqrt{2} }[/tex]
This makes the formula of
CosACosB + SinASinB = Cos(A-B)
Cos(x-[tex]\frac{pi}{6}[/tex]) = [tex]\frac{1}{\sqrt{2} }[/tex]
cos(x- pi/6) = cos(pi/4)
upon writing the general equation we get
x-pi/6 = 2n*pi ± pi/4
x = 2n*pi ± pi/4 -pi/6
so we will have two solutions
x = 2n*pi + pi/4 -pi/6
= 2n*pi + pi/12
and
x = 2n*pi - pi/4 -pi/6
= 2n*pi -5pi/12
Therefore the two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12.
A garden hose fills a 2-gallon bucket in 5 seconds. The number of gallons, g (y), is proportional to the number of seconds, t (x), that the water is running. What is the constant of proportionality?
Answer:
0.4 gallons per second
Explanation:
A function shows the relationship between an independent variable and a dependent variable.
The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.
The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).
Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.
Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second
If the roller at A and the pin at B can support a load upto 4kN and 8kN, repsectively, determine the maximum intensity of the distributed load w, measured in kN/m, so that failure of the supports does not occur.
Solution :
Taken moment about point B;
[tex]$\sum M_B = 0$[/tex]
[tex]$\left[(w)(4)(2)-N_A \sin 30^\circ (3)(\sin 30^\circ)-N_A \cos 30^\circ(3 \cos 30^\circ +4) \right] =0$[/tex]
[tex]$N_A = 1.2376w$[/tex]
[tex]$\sum F_x = 0$[/tex]
[tex]$N_A \sin 30^\circ - B_x = 0$[/tex]
[tex]$B_x = 1.2376w \times \sin 30^\circ$[/tex]
[tex]$B_x = 0.6188w$[/tex]
[tex]$\sum F_y = 0$[/tex]
[tex]$B_y+N_A \cos^\circ - (w \times 4) = 0$[/tex]
[tex]$B_y+(1.2376w) \cos 30^\circ - (w \times 4)=0 $[/tex]
[tex]$B_y = 2.9282w$[/tex]
Now calculating the resultant force at B,
[tex]$F_B= \sqrt{B_x^2+B_y^2}$[/tex]
[tex]$F_B= \sqrt{(0.6188w)^2+(2.9282w)^2}$[/tex]
= 2.9929w
For no failure,
[tex]$N_A<4 \ kN $[/tex]
2.9929 w < 4 kN
[tex]$w < 3.232 \ kN/m$[/tex]
And,
[tex]$F_B < 8 \ kN$[/tex]
2.9929 w < 8 kN
w < 2.673 kN/m
For the safe operation, w = 2.673 kN/m