A) Total Resistance<br> b) Total Current<br> c) Current and Voltage through each resistor

Explain briefly the power analogy between the fluid “water” and the Word of God.

melomori [17]

Throughout the Word of God, water is a symbol, a "type," of the spirit of God, and is ... to represent holy spirit, that is, the divine nature and power of God. ... The average person can go about 60 days without food before he ... record illustrating the analogous relationship between water and the holy spirit.

3 0

3 years ago

1 import java.util.Scanner; 3 public class EqualityAndRelational { 4 public static void main (String args) args) { int userBonus

Anastaziya [24]

Missing Part of the Question

Complete the expression so that userPoints is assigned with 0 if userBonus is greater than 20 (second branch). Otherwise, userPoints is assigned with 10 (first branch

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

( Your solution goes here)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

Answer;

Replace

( Your solution goes here)

With

if(userBonus>20).

The full program becomes

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

if(userBonus>20)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

7 0

3 years ago

True or False? Early engineers used a trial-anderror approach, rather than mathematical and scientific principles when solving

yarga [219]

Answer:

<h2>True Most Especially in the field of Automotive Engineering</h2>

Explanation:

Normally, before the introduction of vehicle diagnostics when a vehicle, mostly automobile/car break down, one could be the vehicle mechanic would only suspect one or two related faults based on the present working condition of the car, the mechanic would perform some trial and error before he could fix the car.

But in recent times, the introduction of vehicle diagnostics devices and software has changed the order as vehicles can be connected to a computer that will scan and tell what the problem is before a possible fix.

3 0

3 years ago

Calculate the magnitude of the velocity and the θ angular direction of the block and the bullet together when the 50 g bullet mo

almond37 [142]

Answer:

Magnitude of the velocity = 16.82 m/s

Angular direction, θ = 52.41°

Explanation:

As given ,

mass of bullet, m₁= 50g = 0.05 kg

speed of bullet , u₁ = 600 m/s

mass of the block , m₂ = 4 kg

speed of the block before collision , u₂ = 12 m/s

direction , θ = 30°

Now,

Assume that the combined velocity of bullet and block after collision = v

and the direction = θ

Now, from the conservation of momentum in x - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ

where v = final velocity after collision

u₁ = initial velocity of bullet before collision = 0

m₁ = mass of the bullet before collision = 0.05 kg

u₂ = velocity of block before collision = 12 cos(30° )

m₂ = mass of block before collision

m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ

⇒ 0 + 4(6√3) = 4.05 vₓ

⇒24√3 = 4.05 vₓ

⇒vₓ = 10.26 m/s

Now, from the conservation of momentum in y - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) $v_{y}$

where v = final velocity after collision

u₁= initial velocity of bullet before collision = 600

m₁ = mass of the bullet before collision = 0.05 kg

u₂ = velocity of block before collision = 12 sin(30° )

m₂= mass of block before collision

m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 ) $v_{y}$

⇒ 30 + 4(6) = 4.05 $v_{y}$

⇒30 +24 = 4.05 $v_{y}$

⇒54 = 4.05 $v_{y}$

⇒ $v_{y}$ = 13.33 m/s

Now, the magnitude of the velocity = √vₓ² + $v_{y}$ ² = √(10.26)² + (13.33)²

= √105.26 + 177.68

= √282.95 = 16.82

The angular direction, θ = $tan^{-1}$ ( $\frac{v_{y} }{v_{x} }$ ) = $tan^{-1}$ ( $\frac{13.33}{10.26}$ ) = $tan^{-1}$ (1.299) = 52.41°

8 0

3 years ago

Fluid Dynamics: How do I find gauge pressure of an air current at various points around a cylinder?A freestream air current of v

DaniilM [7]

Answer:

The answer is as given in the explanation.

Explanation:

The 1st thing to notice is the assumptions required. Thus as the diameter of the cylinder and the wind tunnel are given such that the difference is of the orders of the magnitude thus the assumptions as given below are validated.

Flow is entirely laminar, there's no boundary layer release.
Flow is streamlined, ie, it follows the geometrical path imposed by the curvature.

By D'alembert's paradox, "The net pressure drag exerted on a circular cylinder that moves in an inviscid fluid of large extent is identically zero".Just in the surface of the cylinder, the velocity profile can be given in the next equation:

$V=2Usin\theta$

And the pressure P on the surface of cylinder is given by Bernoulli's equation along the streamline through that point:

$P=P_{_{\infty }}+\frac{1}{2}\rho U^{2}(1-4sin^{2}\Theta ))$

where P_∞ is Pressure at stagnation point, U is the velocity given, ρ is the density of the fluid (in this case air) and θ is the angle measured from the center of cylinder to the adjacent point where your pressure point will be determine.

7 0

3 years ago

A) Total Resistance b) Total Current c) Current and Voltage through each resistor