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Charra [1.4K]
3 years ago
6

A steel rule can be used to check for

Engineering
1 answer:
MAXImum [283]3 years ago
3 0
I THINK THE ANSWER IS B BUT IM NOT SURE OK BYE
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Since you became discouraged not being able to find a job in the San Diego area, you enlarged the area in which you looked for a
nordsb [41]

Answer:

need points 48986

Explanation:

5 0
3 years ago
A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t
Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

tan\phi=\frac{\bigtriangleup l}{h}

tan\phi=\frac{0.08}{1.38}

tan\phi= 0.05797

For small angle of \phi, tan\phi can take as\phi.

\phi = 0.05797 rad.

Thus, the shear strain is 0.05797 rad.

7 0
3 years ago
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
Dafna11 [192]

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
3 years ago
One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
Jlenok [28]
False i think it would be
3 0
2 years ago
import java.util.Scanner; public class FindSpecialValue { public static void main (String [] args) { Scanner scnr = new Scanner(
Hitman42 [59]

Answer:

Java program explained below

Explanation:

FindSpecialNumber.java

import java.util.Scanner;

public class FindSpecialNumber {

public static void main(String[] args) {

//Declaring variable

int number;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//getting the input entered by the user

System.out.print("Enter a number :");

number = sc.nextInt();

/* Based on user entered number

* check whether it is special number or not

*/

if (number == -99 || number == 0 || number == 44) {

System.out.println("Special Number");

} else {

System.out.println("Not Special Number");

}

}

}

_______________

Output#1:

Enter a number :-99

Special Number

Output#2:

Enter a number :49

Not Special Number

7 0
2 years ago
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