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3 years ago

Do electronics harm our mental physical and social health

Physics

2 answers:

Zarrin [17]3 years ago

8 0

Answer:

Just a little bit. You should not be on electronics all the time. I know some people that are so addicted to there electronics that they won't even hang out with people. If you are at the point where you think about things other than school, friends, and family...you should think about getting off of your electronics and go play with a neighborhood friend outside, or have a sleepover! Your whole life should not count on your phone, iPad, or your laptop.

Explanation:

Hope this helps! No recorces included in this writing.

cupoosta [38]3 years ago

6 0

Answer:

(I will talk in my opinion only)

electronics do harm our mental health but they are also helpful ones like mobile both a boon and a curse

physical health is greatly affected and it's something no need to explain

social ( for me ) my only place to be social and talk people is brainly so for me mobile is very important

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A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall

Alexandra [31]

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=14.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.267 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=1.771 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.267 s - 1.771 s$

Finally:

$t=0.495 s$

4 0

4 years ago

Your teacher's car can go from rest to 25 m/s (≈55 mph) in 10 seconds. The car's velocity changes at a uniform rate. Below is a

Mice21 [21]

If the speed of the car changes uniformly, as shown in the graph, this means that the acceleration is constant. As the acceleration is the derivative of the speed

$a = \frac{\delta v}{\delta t}$

Then, this means that the acceleration is the slope of the equation of the line shown in the graph.

Then, we find the acceleration using the following equation:

$a = \frac{v_f- v_i}{t_2-t_1}\\\\a = \frac{25-0}{10}\\\\a = 2.5\ m/s ^ 2$

Then, the acceleration of the car is $2.5\ m/s^2$

5 0

3 years ago

An atom that has only one electron in its valence electron shell is ______ chemically reactive?

umka2103 [35]

The valence electrons are the electrons in the outermost electron shell of an atom.

Explanation:

The number of electrons in an atom's outermost valence shell governs its bondingbehaviour.

That is why elements whose atoms have the same number of valence electrons are grouped together in the Periodic Table.

Generally, elements in Groups 1, 2, and 13 to 17 tend to react to form a closed shell, corresponding to the electron configuration <span><span>s2</span><span>p6</span></span>.

This tendency is called the octet rule, because the bonded atoms have eight valence electrons.

3 0

4 years ago

Read 2 more answers

Two charges are in the configuration indicated here. The first charge, Q1 = –1.00 μC, sits at the origin. The second charge, Q2

blagie [28]

Answer:

$E_{net} = 3.6 \times 10^4 N/C$