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Ostrovityanka [42]
3 years ago
14

I need answers for both questions. thank u if u help.

Engineering
2 answers:
My name is Ann [436]3 years ago
5 0

Answer:

a. Greater the mass, larger the crater

b. Greater the speed, larger the crater

Explanation:

a. They key to answering both questions is understanding that the crater produced by an object is as a result of the force that the object applies when it hits the ground. If it applies a lot of force, a larger crater is produced.

Force is calculated by multiplying mass and acceleration. This means that if either of these is high, the force will be grater and the crater will therefore be larger as well.

Explicitly said, the greater the mass of an object, the larger the size of the crater that it will create.

b. The greater the speed of an object, the larger the size of the crater that it will create as well.

storchak [24]3 years ago
4 0

Answer:

a. Greater the mass, larger the crater

b. Greater the speed, larger the crater

Explanation:

Hope this will help

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Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its the
Andre45 [30]

Answer:

Explanation:

The density of the unit cell of a material, Iron in this case, has to be approximately equal  with its experimental value of 7.87 g/cm³.

The density d = m/v, so what we need to do is calculate the volume of the unit cell and its mass and perform the calculation.

For a BCC crystal structure the length of the side of the cube is given by:

a = 4r/√3

where a is the atomic radius of Iron

first we will convert this radius to cm since we want the density in g/cm³:

0.124 nm x  1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm

a = 4 x 1.24 x 10⁻⁸ cm /√3 = 2.86 x 10⁻⁸ cm

the volume of the cubic cell is:

v = a³ =  ( 2.86 x 10⁻⁸ cm )³ =2.35 x 10⁻²³ cm³

The mass of iron in the body centered cubic cell is obtained from the mass of the atoms in it:

BCC = 2 atoms / unit cell       ( 1/8 from the 8 corners + 1 in the center)

m = 2 atoms/unit cell x 1 mol/ 6.022 x 10²³ atoms  x 55.85 g/mol

   = 1.85 x 10⁻²² g

Therefore,

d = m/v = 1.85 x 10⁻²² g / 2.35 x 10⁻²³cm³ = 7.88 g/cm³

An excelent agreement which confirms that the density of the BCC unit cell agrees with the experimental value.

4 0
4 years ago
One of the dispersive components of an optical instrument is a 0.750-meter focal length monochromator equipped with the same gra
SVETLANKA909090 [29]

Answer:  (a) 1.11 nm/mm

Explanation:

Monochromator is a device which is used to transmits a delectable band of wavelengths of light wavelengths available at the input.

Assuming focal length monochromator  equipped with a                                    1200-groove/mm grating

The formula for the first-order reciprocal linear dispersion is

D^{-1}  = d/n*F where F is the focal length and n belongs to order of the first spectra.

D^{-1} = \frac{(1/1200) mm * 10^6 (nm/mm)/}{1 * 0.75 m * (103 mm/m) } = 1.11 nm/mm

Make sure you have used the conversion factors correctly as it will have a major impact on the calculation of the answer.

7 0
3 years ago
Water flows at a rate of 0.011 m3/s in a horizontal pipe whose diameter increases from 6 to 11 cm by an enlargement section. If
lakkis [162]

Answer:

Pressure change in pipe is 766.96 N/m²

Explanation:

The continuity equation is stated below,

AV = Q

A1V1 = A2V2

Where A is the cross-sectional area, V is the velocity and Q is the fluid flow rate.

To calculate the inlet velocity of the pipe,

V1 = Q/A1

V1 = Q/(π x d1²)

d1 is the inlet diameter of the pipe

Substituting values,

V1 = 0.011/(π x 1/4 x 0.06²)

V1 = 3.89 ms-¹

To determine the outlet velocity,

V2 = Q/A2

d2 is the outlet diameter of pipe

V2 = 0.011/(π x 1/4 x 0.11²)

V2 = 1.157 ms-²

Applying Bernoulli's equation for steady flow between the points,

P1/pg + a1(V1²/2g) + z1 + hp = P2/pg + a2(V2²/2g) + z2 + ht + hL

Collecting like terms,

The kinetic energy correction factor, a = a1 = a2

a((V1² - V2²)/2g) - hL = (P2 - P1)/pg

apg((V1² - V2²)/2g) - hLpg = P2 - P1

ap((V1² - V2²)/2) - hLpg = ∆P

p - density of water, g is the acceleration due to gravity and hL is the head loss due to friction in pipe.

Substituting values,

a = 1.05, p = 1000kg/m³, g = 9.81m/s², hL = 0.66m

∆P = (1000)(1.05)((3.89² - 1.157²)/2) - (0.66 x 1000 x 9.81)

∆P = 7241.56 - 6474.6

∆P = 766.96 N/m²

4 0
3 years ago
Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thic
DIA [1.3K]

Answer:

Plane shell which is option b

Explanation:

The temperature in the case of a steady one-dimensional heat conduction through a plane wall is always a linear function of the thickness. for steady state dT/dt = 0

in such case, the temperature gradient dT/dx, the thermal conductivity are all linear function of x.

For a plane wall, the inner temperature is always less than the outside temperature.

6 0
4 years ago
Write a Lottery class that simulates a lottery. The class should have an array of five integers named lotteryNumbers. The constr
Bezzdna [24]

Answer:

Output:-

Enter the five digit lottery number

Enter the digit 1 : 23

Enter the digit 2 : 44

Enter the digit 3 : 43

Enter the digit 4 : 66

Enter the digit 5 : 33

YOU LOSS!!

Computer Generated Lottery Number :

|12|38|47|48|49|

Lottery Number Of user:

|23|33|43|44|66|

Number Of digit matched: 0

Code:-

import java.util.Arrays;

import java.util.Random;

import java.util.Scanner;

public class Lottery {

int[] lotteryNumbers = new int[5];

public int[] getLotteryNumbers() {

return lotteryNumbers;

}

Lottery() {

Random randomVal = new Random();

for (int i = 0; i < lotteryNumbers.length; i++) {

lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);

}

}

int compare(int[] personLottery) {

int count = 0;

Arrays.sort(lotteryNumbers);

Arrays.sort(personLottery);

for (int i = 0; i < lotteryNumbers.length; i++) {

if (lotteryNumbers[i] == personLottery[i]) {

count++;

}

}

return count;

}

public static void main(String[] args) {

int[] personLotteryNum = new int[5];

int matchNum;

Lottery lnum = new Lottery();

Scanner input = new Scanner(System.in);

System.out.println("Enten the five digit lottery number");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.println("Enter the digit " + (i + 1) + " :");

personLotteryNum[i] = input.nextInt();

}

matchNum = lnum.compare(personLotteryNum);

if (matchNum == 5)

System.out.println("YOU WIN!!");

else

System.out.println("YOU LOSS!!");

System.out.println("Computer Generated Lottery Number :");

System.out.print("|");

for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {

System.out.print(lnum.getLotteryNumbers()[i] + "|");

}

System.out.println("\n\nLottery Number Of user:");

System.out.print("|");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.print(personLotteryNum[i] + "|");

}

System.out.println();

System.out.println("Number Of digit matched: " + matchNum);

}

}

Explanation:

5 0
3 years ago
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