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3 years ago

12

You need to lower your lift onto the mechanical load-holding devices to provide structural support before working under the lift

. True or false?

1 answer:

MAXImum [283]3 years ago

4 0

In my opinion its true

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A worker's hammer is accidentally dropped from the 20th floor of a building under construction. With what velocity does it strik

Illusion [34]

Answer:

Final Velocity (Vf)= 139.864 ft/s

Time (t)= 4,34 s

Explanation:

This is a free fall problem, to solve it we will apply free fall concepts:

In a free fall the acceletarion is gravity (g) = 9,81 m/s2, if we convert it to ft/s^2 = g= 32.174 ft/s^2

Final velocity is Vf= Vo+ g*t[tex]Vf^{2} = Vo^{2} +2*g*h

where h is height (304 ft in this case).

Vo =0 since the hammer wasn't moving when it stared to fall

Then Vf^2= 0 + 2* 32.174 ft/s^2 *304 ft

Vf^2= 19,561.8224 ft^2/s^2

Vf=[sqrt{19561.8224 ft^2/s^2}

Vf=139.864 ft/s

Time t= (Vf-Vo)/g => (139.864 ft/s-0)/32.174 ft/s^2 = 4.34 sec

Good luck!

8 0

3 years ago

“In a trusting relationship, confidential information is kept confidential.” Explain what the limits to confidentiality are and

Sliva [168]

The limits are dangerous situation. Ex. Suicidal actions/thoughts, anything that may put yourself, the individual, or anyone else at risk.

7 0

3 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

4 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Ilya [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

8 0

3 years ago

Technician A says that much of the lighter GMAW welding work in a body shop can be done with a 115-volt welder. Technician B say

Alina [70]

Answer:

c. Both Technicians A and B

8 0

3 years ago