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3 years ago

An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n)

Physics

2 answers:

AVprozaik [17]3 years ago

5 0

That atom is called a ion.

(EYE-on) Like "I've got my ion you.".

maria [59]3 years ago

4 0

Answer:

ion

Explanation:

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A person, of weight mg, is standing on a scale in an elevator. What will the

uranmaximum [27]

more than mg

Explanation:

the pressure

7 0

3 years ago

Identify the labeled parts in the figure.

Hitman42 [59]

Answer:

I. a, c, f and h

II. e

III. b, d, g and i

IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

II. Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e

III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

5 0

3 years ago

A horse walks along a straight path. it travels 3 in one second, 4 m in the next second, and 5 m in the third second. describe t

Temka [501]

Answer: The horse is moving with a uniform acceleration

Explanation:

According to the described situation, we are dealing with a <u>constant acceleration</u> (also called <u>uniform acceleration</u>), since the horse's velocity is changingn by a constant rate.

Let's prove it:

Firstly we are told the horse follows a straight path. In addition we are given its velocity:

3 m/s, then 4 m/s and 5 m/s

Since acceleration $a$ is defined as the change of velocity in time we can calculate it:

$a\frac{4 m/s-3 m/s}{1 s}=1m/s^{2}$ This is the horse's constant acceleration

3 0

3 years ago

An object that is dropped from a height H falls with a constant acceleration of g. The final

GuDViN [60]

The height need to change by 4 to double the final velocity.

<h3>Final velocity of the object</h3>

The final velocity of an object during a free fall is related to maximum height of fall as given the equation below.

v = √2gh

v² = 2gh

v²/h = 2g

v₁²/h₁ = v₂²/h₂

when v₂ = 2v₁, change in height is calculated as;

h₂ = h₁v₂²/v₁²

h₂ = (h₁ (2v₁)²) / (v₁)²

h₂ = 4h₁v₁² / v₁²

h₂ = 4h₁

Thus, the height need to change by 4 to double the final velocity.

Learn more about final velocity here: brainly.com/question/25905661

#SPJ1

4 0

2 years ago

A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass

saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

$W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)$ (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

$d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m$

Furthermore, the angle α is:

$\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°$

Then, you solve the equation (1) for W and replace the values of all parameters:

$W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J$

The work done by the student is 609.97J

6 0

4 years ago