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3 years ago

Which statement describes how the atoms in a conducting wire are organized?(1 point)

Physics

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

Positively charged nuclei are packed in an organized pattern, and the negatively charged valence electrons flow freely.

Explanation:

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Explain why the moon is always half illuminated and half dark no matter where it is in the lunar cycle.

BlackZzzverrR [31]

The easiest way I know to explain it is this:

-- Take a flashlight and a ball into a dark room.

-- Turn on the flashlight and point it at the ball.

-- Half of the ball is lighted up by the flashlight, and the other half is dark.

-- There is no way you can turn or twist the ball to make more or less
than 50% of it lighted up and more or less than 50% of it dark.

Everything in the solar system ... as long as it's shaped like a ball ... is
half illuminated by the sun and half dark.

7 0

3 years ago

FIRST CORRECT ANSWER GETS BRAINLYIEST AND A THANKS!!!

aksik [14]

Your answer is: theres not a certain age. its when the person decides to act mature.

Hope i helped!:)

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4 years ago

4. A chandelier brightens a ballroom after a waiter moves a switch

Leokris [45]

Answer:

because switch moves after a waiter

Explanation:

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3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: