Physics: Why do you sometimes get shocked by static electricity? Explain how it happens and give an example.

Answers: (a) (b) (c) (d) Explanation:This problem is a good example of vertical motion, and the main equations for this situation are as follows:(1)(2)Where: is the height of the stone at a given time is the initial height of the stone is the initial velocity of the stone is the time is the acceleration due to gravity on EarthNow, for parts (a), (b) and (c) we are specifically talking about free fall (where the main condition is that the initial velocity must be zero ). How do we know this? Because...

Why do you sometimes get shocked by static electricity?

Ответ:

cyrusshakeri4814

11.07.2022

Physics

Answers:

(a) $y(t)=300m-4.9m/s^{2}t^{2}$

(b) t=7.82s

(d) t=7.6s

Explanation:

This problem is a good example of vertical motion, and the main equations for this situation are as follows:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V=V_{o}-gt$ (2)

Where:

is the height of the stone at a given time

$y_{o}=300m$ is the initial height of the stone

$V_{o}$ is the initial velocity of the stone

is the time

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

Now, for parts (a), (b) and (c) we are specifically talking about free fall (where the main condition is that the initial velocity must be zero $V_{o}=0$ ).

How do we know this?

Because we are told "the stone is dropped".

Having this clear, let's start with the calculations:

(a) Find the distance of the stone above ground level at time t:

In order to approach this, we will use equation (1), remembering the condition $V_{o}=0$ :

$y(t)=y_{o}-\frac{1}{2}gt^{2}$ (3)

$y(t)=300m-\frac{1}{2}(9.8m/s^{2})t^{2}$ (4)

$y(t)=300m-4.9m/s^{2}t^{2}$ (5) This is the distance of the stone above ground level at time t

(b) How long does it take the stone to reach the ground?:

In this case, y=0 , because we have to find the time when the stone finally hits the ground.

Rewritting (1) with this condition:

$0=y_{o}-\frac{1}{2}gt^{2}$ (6)

Isolating :

$t=\sqrt{\frac{2y_{o}}{g}}$ (7)

$t=\sqrt{\frac{2(300m)}{9.8m/s^{2}}}$ (8)

Then:

t=7.82s (9)

In this part, rewritting equation (2) will be usefull, remembering $V_{o}=0$ and using the timewe found in (9), which is the time when the stone strikes the ground:

V=-gt (10)