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3 years ago

10 properties of metals?

Engineering

2 answers:

Mrrafil [7]3 years ago

6 0

Answer:

1. they are malleable

2. they are ductile

3. they are good conductor of electricity and heat

4. they are solid at room temperature except mercury

5. they are sonorous

6. they are hard

7. they can be hammered into small sheets

8. they are lustrous

9. they are strong

10. they have high density

Misha Larkins [42]3 years ago

4 0

Answer:

Physical Properties of Metals:

Metals can be hammered into thin sheets. ...

Metals are ductile. ...

Metals are a good conductor of heat and electricity.

Metals are lustrous which means they have a shiny appearance.

Metals have high tensile strength. ...

Metals are sonorous. ...

Metals are hard.

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Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

pochemuha

Answer:

import java.util.Scanner;

public class PeopleWeights {

public static void main(String[] args) {

Scanner reader = new Scanner(System.in);

double weightOne = reader.nextDouble();

System.out.println("Enter 1st weight:");

double weightTwo = reader.nextDouble();

System.out.println("Enter 2nd weight :");

double weightThree = reader.nextDouble();

System.out.println("Enter 3rd weight :");

double weightFour = reader.nextDouble();

System.out.println("Enter 4th weight :");

double weightFive = reader.nextDouble();

System.out.println("Enter 5th weight :");

double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

double[] MyArr = new double[5];

MyArr[0] = weightOne;

MyArr[1] = weightTwo;

MyArr[2] = weightThree;

MyArr[3] = weightFour;

MyArr[4] = weightFive;

System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

double average = sum / 5;

System.out.println();

System.out.println("Total weight: " + sum);

System.out.println("Average weight: " + average);

double max = MyArr[0];

for (int counter = 1; counter < MyArr.length; counter++){

if (MyArr[counter] > max){

max = MyArr[counter];

}

System.out.println("Max weight: " + max);

}

import java.util.Scanner;

public class PeopleWeights {

public static void main(String[] args) {

Scanner reader = new Scanner(System.in);

double weightOne = reader.nextDouble();

System.out.println("Enter 1st weight:");

double weightTwo = reader.nextDouble();

System.out.println("Enter 2nd weight :");

double weightThree = reader.nextDouble();

System.out.println("Enter 3rd weight :");

double weightFour = reader.nextDouble();

System.out.println("Enter 4th weight :");

double weightFive = reader.nextDouble();

System.out.println("Enter 5th weight :");

double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

double[] MyArr = new double[5];

MyArr[0] = weightOne;

MyArr[1] = weightTwo;

MyArr[2] = weightThree;

MyArr[3] = weightFour;

MyArr[4] = weightFive;

System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

double average = sum / 5;

System.out.println();

System.out.println("Total weight: " + sum);

System.out.println("Average weight: " + average);

double max = MyArr[0];

for (int counter = 1; counter < MyArr.length; counter++){

if (MyArr[counter] > max){

max = MyArr[counter];

}

System.out.println("Max weight: " + max);

}

8 0

3 years ago

Read 2 more answers

Why is a universal joint useful?

zubka84 [21]

Answer:

the are useful because they transmit a rotary motion they are used in shafts like the driveshaft in a car

Explanation:

3 0

4 years ago

A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t

Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

$\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa$

The design factor for metals is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}$

The design factor for plastic and ceramics is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}$

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17$

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09$

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125$

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12$

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039$

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26$

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

$DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5$

This material will break.

3 0

3 years ago

End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf

sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = ( $\frac{1}{12}$ ) × m × L² ................2

I = ( $\frac{1}{12}$ ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

total force at A will be

Fa = $\sqrt{a(x)^2+a(y)^2}$ ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$

Fa = 57.32 N

3 0

4 years ago

Tires can be recycled instead of thrown out.<br> True<br> False

Arisa [49]

Answer:

True :)

Explanation:

You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.

Hope it helped have a nice day! :)

8 0

3 years ago

10 properties of metals? ​

10 properties of metals?