2=333=3= im single text in comment

Kamila [148]

Answer:

Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)

Explanation:

says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance

(based on image sent in other post)

5 0

3 years ago

Technician A says that a 12 Volt light bulb that draws 12 amps has a power output of 1 watt. Technician B says that a motor that

Artemon [7]

Answer:

Technician B

Explanation:

Resistance, $R=\frac {V}{I}$ where V is the voltage and I is the current in amps

Therefore, $R=\frac {12}{12}=1 ohm$

Power=VI=12*12=144 W

Therefore, the power is 144 W and resistance is 1 Ohm. This implies that technician A is wrong while technician B is correct

6 0

4 years ago

A 65% efficient turbine receives 2 m^3/s of water from a reservoir. The reservoir water surface is 45 m above the centerline of

laiz [17]

Answer:

$P_{out} = 508.071 kW$

Given:

efficiency of the turbine, $\eta$ = 65% = 0.65

available gross head, $H_{G}$ = 45 m

head loss, $H_{loss}$ = 5 m

Discharge, Q = $2 m^{3}$

Solution:

The nozzle is 100% (say)

Available power at the inlet of the turbine, $P_{inlet}$ is given by:

$P_{inlet} = \rho Qg(H_{G} - H_{loss})$ (1)

where

$\rho$ = density of water = 997 $kg/m^{3}$

acceleration due to gravity, g = $9.8 m^{2}$

Using eqn (1):

$P_{inlet} = 997\times 2\times 9.8(45 - 5) = 781.65 kW$

Also, efficency, $\eta$ is given by:

$\eta = \frac{P_{out}}{P_{inlet}}$

$0.65 = \frac{P_{out}}{781.648\times 1000}$

$P_{out} = 0.65\times 781.648\times 1000 = 508071 W = 508.071 kW$

$P_{out} = 508.071 kW$

3 0

3 years ago

This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance

Flauer [41]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

vector<int> jerseyNumber;

vector<int> rating;

int temp;

for (int i = 1; i <= 5; i++) {

cout << "Enter player " << i

<< "'s jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter player " << i

<< "'s rating: ";

cin >> temp;

rating.push_back(temp);

cout << endl;

}

cout << "ROSTER" << endl;

for (int i = 0; i < 5; i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: " << jerseyNumber.at(i)

<< ", Rating: " << rating.at(i) << endl;

char option;

'

while (true) {

cout << "MENU" << endl;

cout << "a - Add player" << endl;

cout << "d - Remove player" << endl;

cout << "u - Update player rating" << endl;

cout << "r - Output players above a rating"

<< endl;

cout << "o - Output roster" << endl;

cout << "q - Quit" << endl << endl;

cout << "Choose an option: ";

cin >> option;

switch (option) {

case 'a':

case 'A':

cout << "Enter a new player's"

<< "jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter the player's rating: ";

cin >> temp;

rating.push_back(temp);

break;

case 'd':

case 'D':

cout << "Enter a jersey number: ";

cin >> temp;

int i;

for (i = 0; i < jerseyNumber.size();

i++) {

if (jerseyNumber.at(i) == temp) {

jerseyNumber.erase(

jerseyNumber.begin() + i);

rating.erase(rating.begin() + i);

break;

}

break;

case 'u':

case 'U':

cout << "Enter a jersey number: ";

cin >> temp;

for (int i = 0; i < jerseyNumber.size();

i++) {

if (jerseyNumber.at(i) == temp) {

cout << "Enter a new rating "

<< "for player: ";

cin >> temp;

rating.at(i) = temp;

break;

}

break;

case 'r':

case 'R':

cout << "Enter a rating: ";

cin >> temp;

cout << "\nABOVE " << temp << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

if (rating.at(i) > temp)

cout << "Player " << i + 1

<< " -- "

<< "Jersey number: "

<< jerseyNumber.at(i)

<< ", Rating: "

<< rating.at(i) << endl;

break;

case 'o':

case 'O':

cout << "ROSTER" << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: "

<< jerseyNumber.at(i) << ", Rating: "

<< rating.at(i) << endl;

break;

case 'q':

return 0;

default:

cout << "Invalid menu option."

<< " Try again." << endl;

}

Explanation:

4 0

3 years ago

Given the potential field in cylindrical coordinates, V = 100/ (z2+1) rho cos φV, and point P at rho = 3m, φ = 60°, z = 2m, find

Margaret [11]

Answer:

V = 30 V

vector (E) = -10*a_p + 17.32*a_Q - 24*a_z

mag(E) = 31.24 V/m

dV / dN = 31.2 V /m

a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z

p_v = 276 pC / m^3

Explanation:

Given:

- The Volt Potential in cylindrical coordinate system is given as:

- The point P is at p = 3 , Q = 60 , z = 2

Find:

values at P for:

a.) V

b.) E

c.) E

d.) dV/dN

e.) aN

f.) rhov in free space

Solution:

a)

Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:

$V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V$

b)

To compute the Electric field from Volt potential we have the following relation:

E = - ∀.V

Where, ∀ is a del function which denoted:

∀ = $\frac{d}{dp}.a_p + \frac{d}{p*dQ}*a_Q + \frac{d}{dz}*a_z$

Hence,

$E = \frac{100*cos(Q)}{z^2 + 1}.a_p+ \frac{100*sin(Q)}{z^2 + 1}.a_Q + \frac{-200*p*zcos(Q)}{(z^2 + 1)^2}.a_z$

Plug in the values for point P:

$E = \frac{100*cos(60)}{5}.a_p+ \frac{100*sin(60)}{5}.a_Q + \frac{-200*3*2*cos(60)}{(5)^2}.a_z\\\\E = - 10*a_p +17.32 *a_Q-24*a_z$

c)

The magnitude of the Electric Field is given by:

E = √((-10)^2 + (17.32)^2 + (24)^2)

E = √975.9824

E = 31.241 N/C

d)

The dV/dN is the Field Strength E in normal to the surface with vector N given by:

dV / dN = | - ∀.V |

dV / dN = | E | = 31.2 N/C

e)

a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:

$a_N = - vector(E) / (dV/dN)\\\\a_N = 10 /31.2 *a_p - 17.32/31.2 *a_Q + 24/31.2 *a_z\\\\a_N = 0.32 *a_p - 0.55 *a_Q + 0.77 *a_z$

f)

The charge density in free space:

p_v = E.∈_o

Where, ∈_o is the permittivity of free space = 8.85*10^-12

p_v = 31.24.8.85*10^-12

p_v = 276 pC / m^3

4 0

3 years ago

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