Which of the circuit diagrams shown in Figure 21-1A is a parallel circuit?

.3.3: Populating a vector with a for loop. Write a for loop to populate vector userGuesses with NUM_GUESSES integers. Read integ

Reptile [31]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

const int NUM_GUESSES = 3;

vector<int> userGuesses(NUM_GUESSES);

int i = 0;

int uGuess = 0;

for(i = 0; i <= userGuesses.size() - 1; i++){

cin >> uGuess;

userGuesses.at(i) = uGuess;

}

cout << endl;

return 0;

}

Explanation:

First inbuilt library were imported. Then inside the main( ) function, 3 was assigned to NUM_GUESSES meaning the user is to guess 3 numbers. Next, a vector was defined with a size of NUM_GUESSES.

Then a for-loop is use to receive user guess via cin and each guess is assigned to the vector.

8 0

3 years ago

The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration

a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>

Forces along x axis direction are as follows

4N along +x axis, so it's taken as +4 N
2N along -x axis , so it's taken as -2N.

Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>

As per Newton's second law, Force = mass × acceleration of the object
Force along x axis= mass × acceleration along x axis= 2N
Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

3 0

2 years ago

Assume that a vaulter is able to carry a vaulting pole while running as fast

rewona [7]

A,walls
Speleleelelelekeke

8 0

3 years ago

Read 2 more answers

Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, t

makvit [3.9K]

Answer:

A) 30 s, 792 m

B) 10.28 s, 4108.2 m = 4.11 km

Explanation:

A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

Using the equations of motion.

v = u + at

v = final velocity = 120 km/h

u = initial velocity = 70 km/h

a = acceleration = 6000 km/h²

t = ?

120 = 70 + 6000t

6000t = 50

t = (50/6000) = 0.0083333333 hours = 30 seconds.

Using the equations of motion further,

v² = u² + 2ax

where x = horizontal distance covered by the car during this time

120² = 70² + 2×6000×x

12000x = 120² - 70² = 9500

x = (9500/12000) = 0.79167 km = 791.67 m = 792 m

B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

For the bullet A

u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

t = T

y = 600T - 4.9T²

At the point where the two bullets pass each other, the vertical heights covered are equal

y = y

450t - 4.9t² = 600T - 4.9T²

But, note that, since T starts reading, 3 seconds after t started reading,

T = (t - 3) s

450t - 4.9t² = 600T - 4.9T²

450t - 4.9t² = 600(t-3) - 4.9(t-3)²

450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)

450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1

600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0

179.4t - 1844.1 = 0

t = (1844.1/179.4) = 10.28 s

Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.

y = 450t - 4.9t²

= (450×10.28) - (4.9×10.28×10.28)

= 4,108.2 m = 4.11 km

Hope this Helps!!!!

6 0

3 years ago

Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha

Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = $\frac{Qa + Qb}{2}$

charge on each sphere = $\frac{-6 + 2}{2}$ = $\frac{-4}{2}$ = <em>-2 units of charge</em>

8 0

3 years ago