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2 years ago

I NEED HELP ASAP WILL AWARD BRAINLIEST

Engineering

2 answers:

CaHeK987 [17]2 years ago

6 0

Answer:

It was a power supply of a circuit that was horizontal of the ladder where they controlled the circuit.

Explanation:

quester [9]2 years ago

3 0

Going CLOCKWISE starting from the LARGEST component: charge (in this case, it’s the battery), switch (that thing is literally called a switch), load (in this case, it’s a light bulb), wire (it’s literally called a wire) i hope that helps!!

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9966 [12]

Answer:

Could ask a family member to help

Explanation:

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3 years ago

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Phil wants to make a dark themed superhero movie. What could be his target demographic?

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male, aged 18-25

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2 years ago

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An equation used to evaluate vacuum filtration is Q = ΔpA2 α(VRw + ARf) , Where Q ≐ L3/T is the filtrate volume flow rate, Δp ≐

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Explanation:

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2 years ago

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.

saul85 [17]

Answer:

a. $L_o$ = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, $L_o$ = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

For the school, $K_{LL}$ = 2

Therefore, we have;

$L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf$

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, $W_d$ = b × $W_{D + L}$ =

∴ $W_d$ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, $W_d$ = 812.8 ft./lb

d. For the uniformly distributed load, we have;

$V_{max}$ = 812.8 × 26/2 = 10566.4 lbs

$M_{max}$ = 812.8 × 26²/8 = 68,681.6 ft-lbs

$v_{max}$ = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

$P_L$ = 1000 lb

$W_{D}$ = 20 × 16 = 320 lb/ft.

$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0

2 years ago

(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record

Effectus [21]

Answer:

import java.util.*;

public class Main {

public static void main(String[] args) {

double milesPerGallon = 0;

int totalMiles = 0;

int totalGallons = 0;

double totalMPG = 0;

Scanner input = new Scanner(System.in);

while(true){

System.out.print("Enter the miles driven: ");

int miles = input.nextInt();

if(miles <= 0)

break;

else{

System.out.print("Enter the gallons used: ");

int gallons = input.nextInt();

totalMiles += miles;

totalGallons += gallons;

milesPerGallon = (double) miles/gallons;

totalMPG = (double) totalMiles / totalGallons;

System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.

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2 years ago