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3 years ago

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Engineering

1 answer:

Kamila [148]3 years ago

7 0

Answer:

(b)False

Explanation:

$I_{xy}$ defined as

$I_{xy}$ = $\int \left (x\cdot y\right )dA$

Where x is the distance from centroidal x-axis

y is the distance from centroidal y-axis

dA is the elemental area.

The product of x and y can be positive or negative ,so the value of $I_{xy}$ can be positive as well as negative .

So from the above expressions we can say that the product of $I_{x},I_y$ is different from $I_{xy}$ .

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The human body gets its energy via the combustion of blood sugar (glucose). if all of the chemical bond energy in 10 g of glucos

Gnesinka [82]

Answer:

speed by mass attain is 55.86 m/s

Explanation:

given data

glucose = 10 g

mass = 100 kg

to find out

speed by mass attain

solution

we know glucose have 180 g molecular weight and

that 1 g glucose produce energy = 2816/180 × 10³ J

so here 10 g of glucose produce energy = 1.56 × $10^{5}$ J

so here energy release = 1/2 × mv²

1.56 × $10^{5}$ = 1/2 × (100)v²

v² = 3.12 × 10³

and v = 55.86 m/s

so speed by mass attain is 55.86 m/s

4 0

3 years ago

How can input from multiple individuals improve design solutions for problems that occur because of a natural disaster, such as

Alla [95]

Answer:

Map and avoid high-risk zones.

Build hazard-resistant structures and houses.

Protect and develop hazard buffers (forests, reefs, etc.)

Develop culture of prevention and resilience.

Improve early warning and response systems.

Build institutions, and development policies and plans.

Explanation:

5 0

2 years ago

Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is

GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

public static void main(String[] args) {

final int MAXCHEESE = 10;

String[] names = new String[MAXCHEESE];

double[] prices = new double[MAXCHEESE];

double[] amounts = new double[MAXCHEESE];

// Three Special Cheeses

names[0] = "Humboldt Fog";

prices[0] = 25.00;

names[1] = "Red Hawk";

prices[1] = 40.50;

names[2] = "Teleme";

prices[2] = 17.25;

System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

System.out.println(names[0] + ": $" + prices[0] + " per pound");

System.out.println(names[1] + ": $" + prices[1] + " per pound");

System.out.println(names[2] + ": $" + prices[2] + " per pound");

Random ranGen = new Random(100);

// error at initialising i

// i should be from 0 to MAXCHEESE value

for (int i = 0; i < MAXCHEESE; i++) {

names[i] = "Cheese Type " + (char) ('A' + i);

prices[i] = ranGen.nextInt(1000) / 100.0;

amounts[i] = 0;

System.out.println(names[i] + ": $" + prices[i] + " per pound");

}

7 0

3 years ago

we want to make a schottky diode on one surface of an n-type semiconductor, and an ohmic contact on the other side. the electron

RoseWind [281]

Answer:

<h2>hope it helps you see the attachment for further information .....✌✌✌✌✌</h2>

7 0

3 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago