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3 years ago

The addition of electron shells results in _____

Physics

1 answer:

vodka [1.7K]3 years ago

6 0

Answer:

shielding of electrons

Explanation:

The addition of electrons shells results in the shielding of electrons away from the nucleus, most importantly the nuclear pull resulting from the charge.

The nucleus pulls electrons to itself due to the net positive charge on it.
As more electronic shell is added, the effect of the pull weakens outward.
The inner shell experiences the nuclear pull more than the outer shell electrons.

The effect is responsible for a wide range of properties of elements.

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during a race, a sprinter increased from 5.0 m/s to 7.5 m/s over a period of 1.25 seconds. what is the sprinters acerage acceler

Sonbull [250]

The answer would be <span>2 m/s/s</span>

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3 years ago

Two identical ions are each missing two electrons, each ion has a charge of 2e. What is the magnitude of the force between the i

xz_007 [3.2K]

Answer:

|F| = 2.09 × 10⁻⁸ assuming that the two ions are point charges.

Explanation:

What's the charge on each ion?

The symbol $e$ here stands for fundamental charge. Each electron carries a negative fundamental charge of -e. Each proton carry a positive fundamental charge of +e.

Molecules and atoms are neutral. They contain an equal number of electrons and protons. Remove one electron from a molecule or atom, and that particle will end up with more protons (which are positive) than electrons. That particle will carry a positive charge of +e become an ion (a cation to be precise.) Remove another electron and the ion will carry a charge of +2e.

For each ion $q = +2 \;e = 2\times 1.60\times 10^{-19}\;\text{C} = 3.2\times 10^{-19}\;\text{C}$ .

What's the size of the electrostatic force between the two ions?

Consider Coulomb's Law for the electrostatic force $F$ between two point charges:

$\displaystyle F = -\frac{k\cdot q_1\cdot q_2}{r^{2}}$ ,

where

$k$ is Coulomb's constant,
$q_1$ and $q_2$ are the charge on the two point charges, and
$r$ is the separation between the two charges.

Make sure that all values are in SI units. Assume that the two ions are small enough that they act like point charges:

$\displaystyle \begin{aligned}F &= -\frac{k\cdot q_1 \cdot q_2}{r^{2}}\\&=-\frac{8.99\times 10^{9}\cdot(3.2\times 10^{-19}) \cdot (3.2\times 10^{-19})}{(2.1\times 10^{-10})^{2}}\\ &= -2.09\times 10^{-8}\;\text{N}\end{aligned}$ .

The value of $F$ is negative, meaning that the two charges will repel each other because they are both positive. The question is asking for the magnitude of this force. Thus drop the sign in front of $F$ to obtain $2.09\times 10^{-8}\;\text{N}$ , which is the magnitude of $F$ .

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3 years ago

Which component of the earth’s atmosphere is decreased due to photosynthesis?

Lana71 [14]

The component of Earth's atmosphere that is decreased due to photosynthesis is carbon dioxide (D).

mark me brainliest!!!!!

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3 years ago

A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck The puck travels for

Norma-Jean [14]

Answer:

Explanation:

Let's analyze the situation presented in order to know which answer is correct.

When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.

Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.

The correct answer is:

C) Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck

4 0

3 years ago

8. Two positive charges (+8.0 mC and +2.0 mC) are separated by 300 m. A third charge is placed at distance r from the +8.0 mC ch

muminat

Answer:

0.20 km

Explanation:

The computation of the distance r is shown below:

This can be calculated by using the following formula

As we know that

$\frac{k_q_1}{r^2} = \frac{k_q_1}{(300 - r)^2}$

where,

q_1 is 8 mC

q_2 is 2 mC

Now placing these values

Now

$\frac{k (8 \times 10^{-3c})}{r^2} = \frac{k (2 \times 10^{-3c})}{(300 - r)^2}$

$\frac{m}{r^2} = \frac{1}{(300 - r)^2}$

$\frac{2}{r} = \frac{1}{300 - r}$

r = 600 - 2r

r = 200 m

r = 0.20 km

We simply applied the above formula to determine the distance of r

Hence, the distance r is 0.20 km

4 0

3 years ago