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3 years ago

if a current of 5 amps flows through a resistance of 40 ohms, what is the voltage across that resistor

Engineering

1 answer:

olasank [31]3 years ago

7 0

Answer:

200V

Explanation:

I = 5A

R = 40Ω

V = IR = (5)(40) = 200V

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My name is Ann [436]

Answer:

cellular service/data

Explanation:

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1 year ago

Write an application that reads three integers, adds all three together, computes an average of the three entries, and computes

Natali5045456 [20]

Answer:

//The program prompts user to input three integers and it displays them, adds and gets their average

//begin

public class Test

{

public static void Main()

{

//input intergers

int[] score = new int[3];

int avg,rem,sum = 0;

for(int i=0;i<3;i++)

{

Console.WriteLine("Enter an integer score ");

score[i] = Convert.ToInt32(Console.ReadLine());

sum = sum + score[i];

}

avg = sum/3;

rem = sum%3;

Console.WriteLine("The average of "+score[0]+","+score[1]+","+score[2]+" is "+avg +" with a remainder of "+rem);

}

3 0

3 years ago

A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

$W_{f}=\gamma(L)(B)(D)$

Where $W_{f}$ is the weight of footing, γ is the unit weight of concrete, L is the length of footing is the width of footing, and D is the depth of footing

Substitute $2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3}$ for γ in the equation

$\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}$

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

$\sigma_{z p}^{\prime}=\gamma(D+B)-u$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0$ for u in the equation

$\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}$

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

$\sigma_{z D}^{\prime}=\gamma D$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3}\) for $\gamma, 0.5 m$ for $D$ and 0 for \(u$ in the equation.

$\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}$

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

$I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}$

Here $I_{e p}$ is the influence factor at the midpoint of soil layer $\sigma_{z^{p}}^{\prime}$ is initial vertical stress, $\sigma_{z^{p}}^{\prime}$ is vertical effective stress, and $Q$ is bearing pressure

Substitute $36 kPa for $\sigma_{z p}^{\prime}, 228.47$ kPa for $q,$ and 9 kPa for $\sigma_{z D}^{\prime}$$ in the equation $\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}$

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0

3 years ago

For the speed equation along centerline of a diffuser, calculate the fluid acceleration along the diffuser centerline as a funct

Marrrta [24]

Answer:

$a = v\cdot \frac{dv}{dx}$ , $v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$ , $\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

Explanation:

Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:

$\dot m_{in} - \dot m_{out} = 0$

$\dot m_{in} = \dot m_{out}$

$\dot V_{in} = \dot V_{out}$

$v_{in} \cdot A_{in} = v_{out}\cdot A_{out}$

The following relation are found:

$\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}$

The new relationship is determined by means of linear interpolation:

$A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x$

$\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x$

After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:

$\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x$

$v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}$

$v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$

The acceleration can be calculated by using the following derivative:

$a = v\cdot \frac{dv}{dx}$

The derivative of the velocity in terms of position is:

$\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.

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3 years ago

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How do you get A's in school

soldier1979 [14.2K]

By being attentive in classes, taking good notes and making sure to put time fourth into your grades.

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3 years ago

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