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4 years ago

How many electrons do all halogen atoms need to fill their outer energy level?

Physics

2 answers:

amm18124 years ago

4 0

Answer:

Seven

Explanation:

Because all halogens have seven valence electrons, they are “eager” to gain one more electron. Doing so gives them a full outer energy level, which is the most stable arrangement of electrons. Halogens often combine with alkali metals in group 1 of the periodic table.

STatiana [176]4 years ago

3 0

Answer:

Halogens only need one electron to complete their valkance shells.

Explanation:

Halogens are present in the VIIA group of the periodic table so seven electrons are already present. one is needed to become stable.

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Imagine you have a collection of identical flat-bottomed coffee filters that can be nested (stacked inside of each other) so tha

skad [1K]

Answer:

the terminal velocity of 14 nested coffee filters is 3.2 m/s

Explanation:

Given the data in the question;

we know that;

The terminal velocity is proportional to the square root of weight.

v ∝ √W

v = k√W

the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged

v/√W = constant

v₂/√W₂ = v₁/√W₁

v₂ = v₁√(W₂ / W₁ )

given that;

v₁ = 0.856 m/s,

W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter

so we substitute

v₂ = 0.856 √(14W₁ / W₁ )

v₂ = 0.856 √( 14( W₁/W₁)

v₂ = 0.856 √( 14(1)

v₂ = 0.856 √( 14 )

v₂ = 0.856 × 3.741657

v₂ = 3.2 m/s

Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s

6 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$