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labwork [276]
3 years ago
11

Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloo

n and wall. Does the balloon move, if so which way
Physics
1 answer:
natka813 [3]3 years ago
3 0

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

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Suppose you have a 100 mhz cpu and a a/d converter with an 8 khz sampling rate. Samples of the a/d converter are 32-bits
seropon [69]

The sampling rate that's given regarding the CPU will produce a rate of 256kps.

<h3>How to calculate the rate? </h3>

The following can be gotten from the question:

Sampling rate = 8KHx

Number of bits = 32

The data rate will be:

= 32 × 8kbps

= 256kps

In conclusion, the sampling rate that's given regarding the CPU will produce a rate of 256kps.

Learn more about rates on:

brainly.com/question/2375289

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2 years ago
If you put 2 eggs in water for 5 secs will it explode???
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3 0
3 years ago
Read 2 more answers
How are gravity and air friction related to a raindrop’s terminal velocity?
Monica [59]
<h2>Answer:</h2>

To answer this question, let's start by clarifying that every body or object that  freely falls in a fluid (in this case the air) <u>has a terminal velocity.</u>

In the particular case of a raindrop (which we assume in a <u>spherical shape</u>) that begins its fall from a certain altitude; it will accelerate (with <u>gravity acceleration</u>) until reaching the terminal velocity, just at the moment when the air friction compensates its weight.

To understand it better, when a raindrop falls, two forces act on it:

1. The force of air friction, also called "drag force" D:

D={C}_{d}\frac{\rho V^{2}}{2}A

Where:

C_{d} is the drag coefficient

\rho is the air density  

V is the velocity

A is the frontal or transversal area of the object (the raindrop)

So, this force is proportional to the transversal area of ​​the falling element (the raindrop) and to the square of the velocity.

2. Its weight due to the gravity force W:

W=m.g

Where:

m is the mass of the raindrop

g is the acceleration due gravity

Then, at the moment when the drag force equals the gravity force:

D=W

The raindrop will have its terminal velocity.

This also means, the larger the raindrop size, the higher its terminal velocity.

In other words, as the drop falls, gravity force pulls it down all the time, but along the descent a force in the opposite direction - upwards - acquires importance, due to the air friction.

6 0
3 years ago
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