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3 years ago

Why is it important foe scientists to know how many valence electrons a given element has

1 answer:

3 0

Because the number of valence electrons of an element determines the properties and in particular the reactivity of that element.

In fact, elements of the first group (i.e. only one valence electron) have high reactivity, because they can easily give away their valence electron to atoms of other elements forming bonds. On the contrary, elements of the 8th group (noble gases) have their outermost shell completely filled with electrons, so they do not have valence electrons, and they have little or no reactivity at all.

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A marble is thrown horizontally with a speed of 7 m/s. When the marble lands, it hastraveled a horizontal displacement of 30 m.

Alex Ar [27]

Answer:

t = 4.28 s

Explanation:

The horizontal speed of a marble = 7 m/s

The horizontal displacement covered by the marble = 30 m

We need to find the time for which the marble is in air. Let the time is t. Using the formula for speed to find it as follows :

$s=\dfrac{d}{t}\\\\t=\dfrac{d}{s}\\\\t=\dfrac{30\ m}{7\ m/s}\\\\=4.28\ s$

So, it will in the air for 4.28 s.

6 0

3 years ago

Cells that help protect the body from disease are the

mash [69]

The answer is either A or D

4 0

3 years ago

Read 2 more answers

A convex mirror of focal length 34 cm forms an image of a soda bottle at a distance of 18 cm behind the mirror. The height of th

Mademuasel [1]

<h2>Answers:</h2>

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

<h2 /><h2>a) Position of the soda bottle (the object) </h2><h2 />

The Mirror equation is:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ (1)

Where:

$f$ is the focal distance

$u$ is the distance between the object and the mirror

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $v$ , let's find $u$ from (1):

$u=\frac{v.f}{v-f}$ (2)

Taking into account the explanation at the beginign of this asnswer:

$f=-34cm$ and $v=-18cm$

The <u>negative signs</u> indicate the <u>focal distance</u> and the <u>distance between the image and the mirror are virtual </u>

Then:

$u=\frac{(-18cm)(-34cm)}{-18cm-(-34cm)}$

$u=38.25cm$ (3) >>>>Position of the soda bottle

<h2>b) magnification of the image </h2><h2 />

The magnification $m$ of the image is given by:

$m=-\frac{v}{u}$ (4)

$m=-\frac{(-18cm)}{38.25cm}$

$m=0.47$ (5)>>>the image is 0.47 smaller than the object

<h2>The fact that this value is positive means the image is upright </h2>

<h2>c) Describe the image </h2><h2 />

According to the explanations and results obtained in the prior answers, the correct option is 9:

<h2>The image is <u>virtual, upright, smaller</u> than the object </h2><h2 /><h2>d) Height of the soda bottle (object) </h2>

Another way to find the magnification is by the following formula:

$m=\frac{h_{i}}{h_{o}}$ (6)

Where:

$h_{i}$ is the image height

$h_{o}$ is the object height

We already know the values of $m$ and $h_{i}$ , let's find $h_{o}$ :

$h_{o}=\frac{h_{i}}{m}$ (7)

$h_{o}=\frac{6.8cm}{0.47}$

$h_{o}=14.46cm$ (8) >>>height of thesoda bottle

6 0

3 years ago

What is the average speed of a car that travels 60 meters in 2<br> seconds?

blondinia [14]

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

6 0

3 years ago

Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi

postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

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

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

$\Delta t= -\frac{v_{oy}}{g}$

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

$t_f=2\frac{v_{oy}}{g}$

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

$v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}$

Finally, since $\sin60\°=\frac{\sqrt{3} }{2}$ and $\sin45\°=\frac{\sqrt{2} }{2}$ , we can conclude that:

$t_{fa}=\sqrt{2}\frac{v_o }{g} \\\\t_{fb}=\sqrt{3}\frac{v_o }{g}\\\\\implies t_{fb}>t_{fa}$

In words, the cannonball b spends more time in the air than cannonball a.

5 0

3 years ago