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Aneli [31]
3 years ago
7

2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m

, report whether the flow is subcritical or supercritical. Also determine the alternate depth (for the same specific energy). Also determine the critical depth.
Engineering
1 answer:
Daniel [21]3 years ago
4 0

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

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________ delivers the precise number of parts to be assembled into a finished product at precisely the right time.
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Answer:

JIT

Explanation:

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6 0
2 years ago
List all information for each order item. Include an item total, which can be calculated by multiplying the Quantity and Paid ea
Zigmanuir [339]

Answer:

#include<iostream>

Using namespace std;

int main()

{

int n, qty;

double price, amount;

cout<<"Number of items ";

cin>>n;

cout<<"ITEM<<"\t"<<"QUANTITY"<<"\t"<<"PRICE"<<"\t"<<"ITEM TOTAL";

for(int i= 1; I<= n; i++)

{

cin>>qty;

cin>>price;

amount = qty * price;

cout<<i<<"\t "<<qty<<"\t"<<price<<"\t"<<amount;

}

}

Explanation

The above program is written in C++ programming language

5 variables are declared and used in the program

n is declared as an integer to represent the total number of items

qty is declared as integer to represent the total quantity of each item

price is declared as double to represent the amount of each individual item

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i is declared as integer to iterated between each items

The amount of each item is calculated within the iteration and also printed immediately

3 0
3 years ago
A sphere is assumed to have the properties of water and has an initial heat generation 46480 W/m^3 How much should the heat gene
Verdich [7]

Answer:

Resulting heat generation, Q = 77.638 kcal/h

Given:

Initial heat generation of the sphere, Q_{Gi} = 46480 W/m^{3}

Maximum temperature, T_{m} = 360 K

Radius of the sphere, r = 0.1 m

Ambient air temperature, T = 25^{\circ}C = 298 K

Solution:

Now, maximum heat generation, Q_{m} is given by:

T_{m} = \frac{Q_{m}r^{2}}{6K} + T                     (1)

where

K = Thermal conductivity of water at T_{m} = 360 K = 0.67 W/m^{\circ}C

Now, using eqn (1):

360 = \frac{Q_{m}\times 0.1^{2}}{6\times 0.67} + 298

Q_{m} = 24924 W/m^{3}

max. heat generation at maintained max. temperature of 360 K is 24924W/m^{3}

For excess heat generation, Q:

Q = (Q_{Gi} - Q_{m})\times volume of sphere, V

where

V = \frac{4}{3}\pi r^{3}

Q = (46480 - 24924)\times \frac{4}{3}\pi\0.1^{3} = 21556\times \frac{4}{3}\pi\0.1^{3} W/m^{3}

Q = 90.294 W

Now, 1 kcal/h = 1.163 W

Therefore,

Q = \frac{90.294}{1.163} = 77.638 kcal/h

3 0
2 years ago
A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a ma
ehidna [41]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

5 0
3 years ago
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