Evaluate, please show work as I don't understand. thanks

Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.

sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh * duration

= 29700 * 6 hours ( 216000 secs )

= 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000 / 185 mi^2

= 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh / volume of runoff in equivalent depth

= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in

5 0

3 years ago

Force = 33 newtons

kicyunya [14]

Answer:

answer

Explanation:

4 0

3 years ago

In this exercise, you will write a Point structure that represents a space in two-dimensional space. This Point should have both

Afina-wow [57]

Answer:

Check the explanation

Explanation:

Points to consider:

We need to take the input from the user

We need to find the manhatan distance and euclidian using the formula

(x1, y1) and (x2, y2) are the two points

Manhattan:

$|x_1 - x_2| + |y_1 - y_2|$

Euclidian Distance:

$\sqrt{(x1 - yl)^2 + (x2 - y2)^2)}$

Code

#include<stdio.h>

#include<math.h>

struct Point{

int x, y;

};

int manhattan(Point A, Point B){

return abs(A.x - B.x) + abs(A.y- B.y);

}

float euclidean(Point A, Point B){

return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));

}

int main(){

struct Point A, B;

printf("Enter x and Y for first point: ");

int x, y;

scanf("%d%d", &x, &y);

A.x = x;

A.y = y;

printf("Enter x and Y for second point: ");

scanf("%d%d", &x, &y);

B.x = x;

B.y = y;

printf("Manhattan Distance: %d\n", manhattan(A, B));

printf("Euclidian Distance: %f\n", euclidean(A, B));

}

Sample output

8 0

3 years ago

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

3 years ago

A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma

MArishka [77]

Answer:

5320.6 Pascal

Explanation:

Manometer is a pressure measuring device use to measure gas pressure .

Pressure difference in Manometer is a function of density,gravity and the height difference of the liquid.

Pressure difference = density x acceleration due to gravity x difference in height of liquid

Density of liquid = specific gravity of object x density of water.

Density of water = 997 kg/m^3

Specific gravity of liquid = 1.7

Density of liquid = 997 x 1.7 =1694.9kg/m^3

g= 9.81 m/s^2

h =320mm = 0.320m

Pressure difference = 1694.9 x 9.81 x 0.320 = 5320.6 Pascal

3 0

4 years ago

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