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3 years ago

Should you ever grab a tool with expose wiring

Engineering

2 answers:

Stolb23 [73]3 years ago

5 0

Answer: No.

Explanation: It could poke your hand and make you bleed or give you a deathly sickness.

Artyom0805 [142]3 years ago

3 0

No! This is common sense!

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(I really need help ASAP please!! this is for science her is the problem)

grandymaker [24]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

Calculate the density of a hydraulic oil in units of kg/m^3 knowing that the density is 1.74 slugs/ft^3. Then, calculate the spe

Amanda [17]

Answer:

Density of oil will be 897.292 kg $m^3$

And specific gravity of oil will be 0.897

Explanation:

We have given density of oil is 1.74 slugs/ $ft^3$

We have to convert this slugs/ $ft^3$ into kg/ $m^3$

We know that 1 slugs = 14.5939 kg

So 1.74 slug = 1.74×14.5939 = 25.3933 kg

And 1 cubic feet = 0.0283 cubic meter

So $1.74slug/ft^3=\frac{1.74\times 14.5939kg}{0.0283m^3}=897.292kg/m^3$

Now we have to calculate specific gravity it is the ratio of density of oil and density of water

We know that density of water = 1000 kg/ $m^3$

So specific gravity of water $=\frac{897.292}{1000}=0.897$

7 0

4 years ago

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stre

velikii [3]

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

The critical yield resolved shear stress is $\sigma = 2.9Mpa$

First we obtain the angle $\lambda$ between the slip direction [121] and [111]

$\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

$\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

$= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

$= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

$\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

Substituting 1 for $u_1$ , 2 for $v_1$ , 1 for $w_1$ , 1 for $u_2$ , 0 for $v_2$ , and 1 for $w_2$

$\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

$\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

$= 54.74 ^o$

The stress is mathematically represented as

$\sigma = \frac{\tau_c}{cos \O cos \lambda }$

$= \frac{2.9}{cos 54.74^o cos 61.87^o}$

$= \frac{2.9}{0.2722}$

$\sigma = 10. 655 MPa$

5 0

3 years ago

Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at l

Tema [17]

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

$m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\$

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

8 0

3 years ago

Exercise 5.46 computes the standard deviation of numbers. This exercise uses a different but equivalent formula to compute the s

ruslelena [56]

Answer:

// This program is written in Java Programming Language

// Comments are used for explanatory purpose

// Program starts here

import java.util.Scanner;

public class STDeviation {

// Declare and Initialise size of Numbers to be 10

int Numsize = 10;

public static void main(String args [] ) {

Scanner scnr = new Scanner(System.in);

// Declare digits as double

double[] digits = new double[Numsize];

System.out.print("Enter " + Numsize + " digits: ");

// Input digits using iteration

for (int i = 0; i < digits.length; i++)

{

digits[i] = scnr.nextDouble();

}

// Calculate and Print Mean/Average

System.out.print("Average: " + mean(digits)+'\n');

// Calculate and Print Standard Deviation

System.out.println("Standard Deviation: " + deviation(digits));

}

// Standard Deviation Module

public static double deviation(double[] x) {

double mean = mean(x);

// Declare and Initialise deviation to 0

double deviation = 0;

// Calculate deviation

for (int i = 0; i < x.length; i++) {

deviation += Math.pow(x[i] - mean, 2);

}

// Calculate length

int len = x.length - 1;

return Math.sqrt(deviation / len);

}

// Mean Module

public static double mean(double[] x) {

// Declare and Initialise total to 0

double total = 0;

// Calculate total

for (int i = 0; i < x.length; i++) {

total += x[i];

}

// Calculate length

int len = x.length;

// Mean = total/length

return total / len;

}

4 0

3 years ago