An electric current in a metal consists of moving

An atom gains an additional electron. What is the overall charge of the ion that is formed?

jeka57 [31]

An is formed when an atom loses or gains one or more electrons. Because the number of electrons in an ion is different from the number of protons, an ion does have an overall electric charge. Consider how a positive ion can form from an atom. The left side of the illustration below represents a sodium (Na) atom

8 0

3 years ago

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El espectro visible en el aire está comprendido entre la longitud de onda 450 nm del color azul, Determina la velocidad de propa

TiliK225 [7]

Answer:

v = 2,99913 10⁸ m / s

Explanation:

The velocity of propagation of a wave is

v = λ f

in the case of an electromagnetic wave in a vacuum the speed that speed of light

v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

λ = λ₀ / n

where n is the index of refraction of the material medium.

Therefore the expression is

v = $\frac{\lambda_o}{n} f$

Let's look for the frequency of blue light in a vacuum

f = $\frac{c }{\lambda_o}$

f = $\frac{3 \ 10^8}{450 \ 10^{-9}}$

f = 6.667 10¹⁴ Hz

the refractive index of air is tabulated

n = 1,00029

let's calculate

v = $\frac{450 \ 10^{-9} }{1.00029} \ 6.667 \ 10^{14}$ 450 10-9 / 1,00029 6,667 1014

v = 2,99913 10⁸ m / s

we can see that the decrease in speed is very small

8 0

2 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$