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coldgirl [10]
3 years ago
9

After reading through the code, what will happen when you click run?* 1 point when run move forward while there is a pile do rem

ove 1 O Nothing will happen. O The farmer will dig a hole and add to the pile of dirt. O The farmer will never stop removing dirt. O The farmer will remove dirt as long as there is a pile, then stop when the pile is done. 1 point After reading through the code, what will happen when you click run? * when an​

Engineering
1 answer:
KATRIN_1 [288]3 years ago
5 0

Explanation:

I'm not exactly a master at coding, but I'm pretty sure that:

The farmer will remove dirt as long as there is a pile, then stop when the pile is done.

You might be interested in
Two plates are separated by a 1/4 in space. The lower plate is stationary; the upper plate moves at 10 ft/s. Oil (viscosity of 2
Montano1993 [528]

Answer:

τ = 0.25 lbf/in²

Explanation:

given that the oil viscosity, μ =  2.415 lb/ft-s

gap between plates = 1/4 inches = 1/4*12 = 1/48 ft

recall from newtons law of viscosity;

shear stress τ = μ du/dy =

  τ = (2.415 lb/ft-s) (10 ft/s)/(1/48) ft

τ = 1159.2 lb/ft-s²

we know that, 1 slug = 32.174 lb

lb = 1/32.174 slug

∴  τ = 1159.2/32.174 slug/ft-s² = 36 slug/ft-s²

τ = 36 slug/ft-s²

multiply both the numerator and denominator by ft, this gives

τ = 36 slug-ft/ft²-s²

τ = 36 lbf/ft² where 1 slug-ft/s² = 1lbf

since 1 ft = 12 inch = 1 ft² = 12² in² = 144 in²

∴  τ = 36/144 lbf /in² = 0.25 lbf/in²

τ = 0.25 lbf/in²

6 0
3 years ago
The solid spindle AB is connected to the hollow sleeve CD by a rigid plate at C. The spindle is composed of steel (Gs = 11.2 x 1
dalvyx [7]

Answer:

T_max = 12.63 kip.in

Ф_a = 1.093°

Explanation:

Given:

- The modulus of rigidity of solid spindle G_ab = 11.2 * 10^6 psi

- The diameter of solid spindle d_ab = 1.75 in

- The allowable stress in solid spindle τ_ab = 12 ksi

- The modulus of rigidity of sleeve G_cd = 5.6 * 10^6 psi

- The outer diameter of sleeve d_cd = 3 in

- The thickness of sleeve t = 0.25

- The allowable stress in sleeve τ_cd = 7 ksi

Find:

- The largest torque T that can be applied to end A that does not exceed allowable stresses and sleeve angle of twist 0.375°

- The corresponding angle through which end A rotates.

Solution:

- Calculate the polar moment of inertia of both spindle AB and sleeve CD.

   Spindle AB:    c_ab = 0.5*d_ab = 0.5(1.75) = 0.875 in

                           J_ab = pi/2 c^4 = pi/2 0.875^4 = 0.92077 in^4

   Sleeve CD:  c_cd1 = 0.5*d_cd = 0.5(3) = 1.5 in , c_cd2 = c_cd1 - t = 1.25

                     J_cd = pi/2 (c_cd1^4 - c_cd2^4)= pi/2(1.5^4-1.25^4) = 4.1172 in^4

- The stress criteria for maximum allowable torque in spindle AB:

                    T_ab = J_ab*τ_ab / c_ab

                    T_ab = 0.92077*12 / 0.875

                    T_ab = 12.63 kip.in

- The stress criteria for maximum allowable torque in sleeve CD:

                    T_cd = J_cd*τ_cd / c_cd1

                    T_cd = 4.1172*7 / 1.5

                    T_cd = 19.21 kip.in

- The angle of twist criteria for point D:

                    T_d = J_cd*G_cd*Ф / L

                    T_d = 4.1172*5.6*10^6*0.006545 / 8

                    T_d = 18.86 kip.in

- The maximum allowable Torque for the structure is:

                    T_max = min ( 12.63 , 19.21 , 18.86 )

                    T_max = 12.63 kip.in

- The angle of twist of end A:

                    Ф_a = Ф_a/d = Ф_a/b + Ф_c/d:

                    T_max* ( L_ab / J_ab*G_ab + L_cd / J_cd*G_cd)

                    12.63*(12/0.92*11.2*10^6 + 8/4.117*5.6*10^6)

                    0.01908 rads = 1.093°

3 0
3 years ago
The speed of a vehicle is reduced with a constant acceleration from 72km/h to 18
Radda [10]

Answer:

The correct answer will be "1477.84 N".

Explanation:

Given that,

Mass,

m = 1.6 mg

or,

   = 1600 kg

Initial velocity,

u = 72 km/h

  = 72\times \frac{5}{18} \ m/s

  = 20 \ m/s

Final velocity,

v = 18 km/h

  = 18\times \frac{5}{18}

  = 5 \ m/s

Covered distance,

s = 250 m

By using the below relation, we get

⇒  v^2=u^2+2as

On putting the values, we get

⇒  (5)^2=(20)^2+2\times a\times 250

⇒      a=-0.75 \ m/s^2 (shows the deceleration)

Slope will be given as 1 in 25, then

⇒  Sin \theta=\frac{1}{25}

           \theta=2.3^{\circ}

hence,

As we know,

⇒  \Sigma F=ma

or,

⇒  Braking \ force+350-mgSin\theta=ma

⇒  Braking \ force=ma+mgSin\theta-350

On substituting all the values, we get

⇒                           =1600(0.75+1600\times 9.81 Sin(2.3^{\circ})-350

⇒                           =1477.84 \ N

7 0
3 years ago
Biologists use a sequence of letters A, C, T, and G to model a genome. A gene isa substring of a genome that starts after a trip
kogti [31]

Answer:

You did not mention the programming language for implementation so i am writing a JAVA code.

import java.util.Scanner; // to get input from user

public class Genome{

public static void main(String[] args) { //start of main() function body

Scanner input = new Scanner(System.in); //creates Scanner object

System.out.print("Enter a genome string: ");

//prompts user to enter a genome string

String genome = input.nextLine();

//reads the input genome string and stores it into genome variable

boolean gene_found = false;

//variable gene_found of boolean type that has two value true or false

int startGene = 0; // stores starting of the gene string

for (int i = 0; i < genome.length() - 2; i++) {

//loop moves through genome string until the third last gene character

String triplet = genome.substring(i, i + 3);

//stores the triplet of genome substring

if (triplet.equals("ATG")) {

//if value in triplet is equal to ATG

startGene = i + 3;

//3 is added to i-th position of the genome string

}

else if (((triplet.equals("TAG")) || (triplet.equals("TAA")) || (triplet.equals("TGA"))) &&(startGene != 0))

//checks if the genome ends with one the given triplets TAG TAA and TGA

{ String gene = genome.substring(startGene, i);

gene stores substring of genome string from startGene to the position i

if (gene.length() % 3 == 0)

//if the the mod of gene length is 0 then the gene is found

{gene_found = true;

System.out.println(gene); //returns the found gene

startGene = 0;} } }

if (!gene_found) //if gene is not found returns the message below

System.out.println("no gene is found"); }  }

Explanation:

This program first asks user to enter a genome string.

The loop starts from the first character of the entered string and this loop continues to execute until the value of i is 2 less than the genome input string length.

triplet variable stores first 3 characters of the genome string in first iteration and then moves through the string taking 3 characters each. This is done by dividing genome string to substring of 3 characters.

If condition checks if the 3 characters of genome string matches ATG using equals() function. If it is true this means start of genome is reached and these triplets are stored in startGene.

Else condition checks the end of the genome as the genome ends before one of TAG, TAA or TGA triplets. So this is checked here.

gene variable holds the triplet value stored in startGene and the value stored in index position i which means it holds the start of the genome till the end of the genome sequence. The end which is pointed by i variable is 2 less than the genome length and it is stored in gene variable.

After the loop ends the substring stored in gene variable is checked for a valid genome sequence by mod operator. If the length of the value stored in gene variable mod 0 is equal to 0 this means genome sequence is found and this string sequence stored in gene is displayed else the no gene is found message is displayed on output screen.

7 0
3 years ago
write a paragraph on how iron and copper(separately) is manufactured, what it consists of and processes that it went through to
trasher [3.6K]

Answer:

Iron is manufactured in a blast furnace. First, iron ore is mixed with coke and heated to form an iron-rich clinker called 'sinter'. Sintering is an important part of the overall process as it reduces waste and provides an efficient raw material for iron making. Coke is produced from carefully selected grades of coal.

Copper is typically extracted from oxide and sulfide ores that contain between 0.5 and 2.0% copper. ... Regardless of the ore type, mined copper ore must first be concentrated to remove gangue or unwanted materials embedded in the ore. The first step in this process is crushing and powdering ore in a ball or rod mill.

5 0
3 years ago
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