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dexar [7]
1 year ago
15

Inches of mercury is something that would only be used in America. Elsewhere in the world, the mercury height would be in millim

eters, or about 760mm (76cm). Mercury has a density of about 13.56 g/cm3, or in other words, each cm of mercury weighs about 13,300 dynes. So 760mm of mercury weighs 1,013,000 dynes. 1 mb is 1 dyne/cm2, so what is the air pressure in mb
Engineering
1 answer:
dusya [7]1 year ago
5 0

Answer:

it alot of mecury but 123443 is the anzswer

Explanation:

there is nun it just listen to god kids

You might be interested in
CHALLENGE ACTIVITY 2.8.1: Using constants in expressions. The cost to ship a package is a flat fee of 75 cents plus 25 cents per
mihalych1998 [28]

Answer:

Weight(lb): 10

Flat fee(cents): 75

Cents per pound: 25

Shipping cost(cents): 325

Explanation:

we run this as a jave programming language

import java.util.Scanner;

public class Shipping Calculator {

   public static void main (String [] args) {

       int shipWeightPounds = 10;

       int shipCostCents = 0;

       final int FLAT_FEE_CENTS = 75;

       

        final int CENTS_PER_POUND = 25;

       shipCostCents = FLAT_FEE_CENTS + CENTS_PER_POUND * shipWeightPound

       /* look up the solutioin above */

       System.out.println("Weight(lb): " + shipWeightPounds);

       System.out.println("Flat fee(cents): " + FLAT_FEE_CENTS);

       System.out.println("Cents per pound: " + CENTS_PER_POUND);

       System.out.println("Shipping cost(cents): " + shipCostCents);

   }

}

7 0
2 years ago
Q10. Select the correct option for the following questions – (10 points, 2 each) a. After an edge dislocation has passed through
Gnom [1K]

Answer:

a. True - Because the atomic arrangements of that region is disorderer because of the extra half plane atoms in between the line

b.Slip

C. Strength theoretical is greater than strength experimental

d. Shear stress

e. Highest linear density

4 0
2 years ago
How does miep build suspense in this passage
siniylev [52]

Then I walked into Mr. and Mrs. Frank's bedroom. On the floor, amidst the chaos of papers and books, my eye lit on the little red-orange checkered, cloth-bound diary that Anne had received from her father on her thirteenth birthday. She loves Anne and respects her work.

I hope this helps you.

3 0
2 years ago
Read 2 more answers
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor ack
Elza [17]

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

3 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
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