Although studs are sometimes spaced 24" O.C. in residential structures, a spacing of____

(1) 1. (15 points/ 3 points each) (a) Draw the binary search tree that is created if the following numbers are inserted in the t

Alex777 [14]

Answer:

The binary search tree BST that is created is shown in the figure in the attached file

The missing part of the question is to draw the balanced binary search tree containing the same numbers given in the question.

6 0

3 years ago

The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbac

bazaltina [42]

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

$\bold{Area =B \times D_f}$

$=6\times 5\\\\=30 \ ft^2$

In point b, Calculating the wetter perimeter.

$\bold{P_w =B+2\times D_f}$

$= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft$

In point c, Calculating the hydraulic radius:

$\bold{R=\frac{A}{P_w}}$

$=\frac{30}{16}\\\\= 1.875 \ ft$

In point d, Calculating the value of Reynolds's number.

$\bold{Re =\frac{4VR}{v}}$

$=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\$

$=750,000 V$

Calculating the velocity:

$V= \sqrt{\frac{8gRS}{f}}$

$= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\$

$\sqrt{f}=\frac{3.108}{V}\\\\$

calculating the Cole-brook-White value:

$\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\$

$\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})$

After calculating the value of V it will give:

$V= 25.18 \ \frac{ft}{s^2}\\$

In point a, Calculating the value of Froude:

$F= \frac{V}{\sqrt{gD}}$

$= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\$

$= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98$

The flow is supercritical because the amount of Froude is greater than 1.

Calculating the channel flow rate.

$Q= AV$

$=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\$

4 0

3 years ago

Global Courier Services will ship your package based on how much it weighs and how far you are sending the package. Packages abo

denis23 [38]

Answer:

The code will be:

#include <stdio.h>

#include <stdlib.h>

main () {

double weight, shippingCharge, rate, segments;

int distance;

printf("Enter the weight: \n");

scanf("%lf", &weight);

printf("Enter the distance: \n");

scanf("%i", &distance);

if (weight <= 10) {

printf("Rate is $3.00 \n");

rate = 3;

} else {

printf("Rate is $5.00 \n");

rate = 5;

}

if (distance % 500 == 0) {

segments = distance / 500;

} else {

segments = distance / 500 + 1;

}

shippingCharge = rate * segments;

if (distance >1000) {

shippingCharge = shippingCharge + 10;

}

printf("Your shipping charge is $%lf\n", shippingCharge);

system ("pause");

}

8 0

3 years ago

In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2

sp2606 [1]

Answer:

$l=24mm$

Explanation:

From the question we are told that:

Plane strain fracture toughness of $T=55 MPa-m1/2$

Y value $Y=1.0$

Stress level of $\sigma =200 MPa$

Generally the equation for length of a surface crack is mathematically given by

$l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2$

$l=\frac{1}{3.142}(\frac{55}{1*200})^2$

$l=0.024m$

Therefore

in mm

$l=24mm$

6 0

3 years ago

2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11

ahrayia [7]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² = 1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² = 9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ = (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵) × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )

0 - 0.6014925 + Rₓ = 0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R $_y$ = m'( V₂sin(60°) - 0.5 )

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R $_y$ = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R $_y$ = 0.092728( 0.305187 )

⇒ 1.5484 + R $_y$ = 0.028299

R $_y$ = 0.028299 - 1.5484

R $_y$ = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R $_y$ ² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

7 0

3 years ago

Although studs are sometimes spaced 24" O.C. in residential structures, a spacing of_____ O.C. is more commonly used.