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3 years ago

What is the magnitude of the charge of an electron?

Engineering

2 answers:

makkiz [27]3 years ago

8 0

Answer: Physical Constant

Explanation:

viktelen [127]3 years ago

7 0

Answer:

The magnitude of the charge of an electron is −1.602×10−19C

I really sorry if it incorrect

hope you have a nice dayy!

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Where can you find the grade for a piece of property?

Nadya [2.5K]

Answer:

Simply put, the grade or grading around your house is the level of the ground. The ground level and how it's graded is the deciding factor of where storm water will flow.

Explanation:

5 0

3 years ago

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

A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (SG 0.9)and rotates about its vertical longitudinal axis

egoroff_w [7]

Answer: $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

Explanation: To determine the pressure of a liquid in a rotating tank,it is used:

p = $\frac{p_{fluid}.w^{2}.r^{2} }{2}$ - γfluid . z + c

where:

$p_{fluid}$ is the liquid's density

w is the angular velocity

r is the radius

γfluid.z is the pressure variation due to centrifugal force.

For this question, the difference between a point on the circumference and a point on the axis will be:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}.r_{B} ^{2} }{2}$ - γfluid. $z_{B}$ - ( $\frac{p_{fluid}.w^{2}.r_{A} ^{2} }{2}$ - γfluid. $z_{A}$ )

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$ - γfluid( $z_{B}$ - $z_{A}$ )

Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$

$p_{B} - p_{A} = \frac{0.9.10^3.40^2}{2}(0.2^2 - 0)$

$p_{B} - p_{A}$ = 28800

The difference in pressure between two points, one on the circumference and the other on the axis is $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

8 0

3 years ago

25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

5 0

3 years ago

A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.

Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

$E_{in}- E_{out}=E_{system,final}- E_{system,initial}$

Where

$E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}$

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

$E_{in}- E_{out}=-Q_{out}+W_{in}$

Regarding the energy system we have the following:

$E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}$

By doing the calculations we have:

$E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ$

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

$W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ$

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0

3 years ago

What is the magnitude of the charge of an electron?​

What is the magnitude of the charge of an electron?