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3 years ago

What does the atomic number of an element represent?

Physics

1 answer:

Zinaida [17]3 years ago

7 0

The number of protons in an atom is represented by the atomic number.

hope this helps

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You look down at your phone GPS and it tells you the local park is 2 miles away. This is not a straight path and you will need t

Dafna11 [192]

Answer:

<h2><u>Distance.</u></h2>

<u> $\infty \infty$ </u>

6 0

3 years ago

Read 2 more answers

How to Know 1 Light Year? Please help me..

max2010maxim [7]

Explanation:

one light year is the distance traveled by the light in 365.25 days. Thus,

1 year is 365.25 days.

1 day is 24 hour

1 hour is 60 minute

1 minute is 60 second

thus, 365.25*24*60*60 is 31557600 second.

Answer:

1 light year is approximately 9.4607 trillion km

8 0

3 years ago

A microscope has an eyepiece with a 1.8 cm focal length and a 0.8 cm focal length objective lens. Assuming a relaxed eye, calcul

igomit [66]

Answer:

$0.848\ \text{cm}$

$232.66$

Explanation:

N = Near point of eye = 25 cm

$f_o$ = Focal length of objective = 0.8 cm

$f_e$ = Focal length of eyepiece = 1.8 cm

l = Distance between the lenses = 16 cm

Object distance is given by

$v_o=l-f_e\\\Rightarrow v_o=16-1.8\\\Rightarrow v_o=14.2\ \text{cm}$

$u_o$ = Object distance for objective

From lens equation we have

$\dfrac{1}{f_o}=\dfrac{1}{u_o}+\dfrac{1}{v_o}\\\Rightarrow u_o=\dfrac{f_ov_o}{v_o-f_o}\\\Rightarrow u_o=\dfrac{0.8\times 14.2}{14.2-0.8}\\\Rightarrow u_o=0.848\ \text{cm}$

The position of the object is $0.848\ \text{cm}$ .

Magnification of eyepiece is

$M_e=\dfrac{N}{f_e}\\\Rightarrow M_e=\dfrac{25}{1.8}\\\Rightarrow M_e=13.89$

Magnification of objective is

$M_o=\dfrac{v_o}{u_o}\\\Rightarrow M_o=\dfrac{14.2}{0.848}\\\Rightarrow M_o=16.75$

Total magnification is given by

$m=M_eM_o\\\Rightarrow m=13.89\times 16.75\\\Rightarrow m=232.66$

The total magnification is $232.66$ .

3 0

3 years ago

What is it called when a neutral object is given a temporary charge?

fgiga [73]

It's called induced charge separation I believe.

4 0

3 years ago

You are in a hot air balloon (yes, another balloon problem!) rising from the ground at a constant velocity of 2.00 m/s upward. t

Daniel [21]

If all you need is the initial speed of the cork, you can solve this using only two of your given:
2.00 m.s upward and 6.60 m.s horizontally.

If you take in consideration the movement of the cork, you know that it was both going up and forward at the same time, this means that it was moving at a diagonal direction. Now you can solve this by using the Pythagorean theorem where:

$c = \sqrt{ a^{2} + b^{2} }$

Why? Because the vertical and the horizontal motion creates a movement that is diagonal, which when put in a free-body diagram, creates a right triangle.

Going back to your problem, when applying this, the diagonal of a right triangle is the hypotenuse, so this is what you are looking for. The horizontal and vertical motion will represent the other 2 sides of the triangle.

Now let's put that into your formula:

$c = \sqrt{ a^{2} + b^{2} }$

$Vi = \sqrt{ Vx^{2} + Vy^{2} }$

Where: Vx is your horizontal velocity
Vy is your vertical velocity
Vi is your initial velocity

Now let's put in your given:

$Vi = \sqrt{ Vx^{2} + Vy^{2} }$
$Vi = \sqrt{ 6.60^2} + 2.00^{2} }$
$Vi = \sqrt{ 43.56 + 4.00 }$
$Vi = \sqrt{ 47.56 }$
$Vi = 6.8964 m/s$

So your initial velocity is 6.8964 m/s or 6.90 m/s

5 0

4 years ago