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2 years ago

Explain why it is not advisable to wear Ornament like ring at work shop

1 answer:

8 0

The reason that it is not advisable to wear Ornament like ring at work shop is that rings can be easily get caught and fingers or the hands of the person can be injured, cut, scared etc.

<h3>Why should you not wear a ring at work?</h3>

Jewelry is known to bring about a lot of safety hazards for people working around chemicals and others.

Note that the reason that it is not advisable to wear Ornament like ring at work shop is that rings can be easily get caught and fingers or the hands of the person can be injured, cut, scared etc.

Learn more about Ornament from

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

2 years ago

Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio

SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0

3 years ago

Why is it important to keep hand tools free of oil and grease?

Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0

3 years ago

Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,

aalyn [17]

Answer:

import java.util.Scanner;

public class InputExample {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int birthMonth;

int birthYear;

birthMonth = scnr.nextInt();

birthYear = scnr.nextInt();

System.out.println(birthMonth+"/"+birthYear);

}

3 0

3 years ago

Explain why it is not advisable to wear Ornament like ring at work shop​

Explain why it is not advisable to wear Ornament like ring at work shop