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jok3333 [9.3K]
3 years ago
11

Write a statement that increases numPeople by 5. Ex: If numPeople is initially 10, the output is: There are 15 people.

Engineering
1 answer:
BartSMP [9]3 years ago
8 0

Answer:

import java.util.Scanner;

public class AssigningNumberToVariable {

public static void main (String [] args) {

int numPeople;

int numPeople_inc;

numPeople = 10;

/* solution */

numPeople_inc=numPeople+5;

System.out.print("There are ");

System.out.print(numPeople_inc);

System.out.println(" people.");

}

}

Explanation:

1. Declare a new variable called: numPeople_inc

2. Add 5 each time: numPeople_inc=numPeople+5;

3. Print the answer:  numPeople_inc=numPeople+5;

Optional, just write the line in the place where says: " your solutions goes here":

numPeople = numPeople+5;

You might be interested in
In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivale
sashaice [31]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivalent of this circuit using phasors.

a. If you write the Thevenin voltage, Vth, in phasor form, what is the magnitude of this phasor? Put your answer in the box below without units.

b. What is the value of the angle associated with the phasor Vth, in degrees?

c. Now, calculate the Thevenin impedance, Zth. What is the magnitude of this phasor?

d. What is the angle associated with the phasor Zth, in degrees?

Answer:

Vth = 6 < 45° V

Zth = 1.414 < 45°

a. The magnitude of the Thevenin voltage is 6 V

b. The phase angle of the Thevenin voltage is 45°

c. The magnitude of the Thevenin impedance is 1.414 V

d. The phase angle of the Thevenin impedance is 45°

Explanation:

The given voltage is

V(t)=12cos(2000t+45)

In phasor form,

V(t) = 12 < 45° V

So the magnitude of voltage is 12 V and the phase angle is 45°

Also the frequency ω = 2000

then the inductance is

L₁ = L₂ = L₃ = jωL = j×2000×0.003 = j6 Ω

and the capacitance is

C₁ = 1/jωC = 1/(j×2000×250x10⁻⁶) = -j2 Ω

and the resistance is

R₁ = R₂ = 2 Ω

Thevenin voltage:

The Thevenin voltage is the voltage that appears across the open-circuited terminals a-b (after removing L₃)

The Thevenin voltage is given by

Vth = V(t) × [ (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ]

Please note that there is no current flow in the capacitor due to open-circuited terminal a-b

Vth = 12 < 45°  × [ (2 + j6) / (2 + j6) + (2 + j6) ]

Vth = 12 < 45°  × [ (2 + j6) / (4 + j12) ]

Vth = 4.24264 + j4.24264 V

In phasor form,

Vth = 6 < 45° V

a. The magnitude of the Thevenin voltage is 6 V

b. The phase angle of the Thevenin voltage is 45°

Thevenin Impedance:

The Thevenin Impedance is the impedance of the circuit calculated when looking from the terminal a-b

Zth = [ (R₁ + L₁) × (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ] + (-j2)

Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2

Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2

Zth = [ (-32 + j24) / (4 + j12) ] -j2

Zth = [ (1 +j3) ] - j2

Zth = 1 + j Ω

In phasor form,

Zth = 1.414 < 45°

c. The magnitude of the Thevenin impedance is 1.414 V

d. The phase angle of the Thevenin impedance is 45°

8 0
3 years ago
Weight limit for a pallet
hammer [34]

Answer:

standard pallet can withstand 4,600lbs

Explanation:

6 0
3 years ago
Read 2 more answers
A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi
11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

h_f=\dfrac{FLV^2}{2gd}

h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}

h_f=40.77m

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

P=5.5\times 10^5\ Pa

P= 5.5 bar

5 0
3 years ago
In a paragraph of 125 words, explain at least three ways that engineers explore possible solutions in their
joja [24]

Answer:

Three ways that engineers explore possible solutions in their projects are;

1) Prototyping

2) Simulation

3) Calculations

Explanation:

1) Prototyping is the process of experimental testing of samples of design, or model of a product with the possibility of the inclusion of control of parameters in order to determine the workability of a solution.

2) Simulation is the process of creating an imitation of a situation, operation or process which can be used to determine if a particular solution will be able to work as required in the simulated environment of a problem.

3) Calculations are used to find preliminary results of particular situations, their cause and effects based on scientific laws, theories and hypothesis such that the factor of the problem is equated with the available ideas to find the best possible solution.

3 0
3 years ago
A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig
Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0
3 years ago
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