1.- La gráfica muestra la variación de la aceleración a de un objeto con el tiempo t.

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible

Leto [7]

Answer:

a) 298.5 nm , 522.4 nm and b) radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

λₙ = λ₀ / n

λₙ = 400 nm in the void

λₙ = 400 / 1.34

λₙ= 298.5 nm

λ₀ = 700 nm

λₙ = 700 / 1.34

λₙ = 522.4 nm

The radiation frequency does not change

4 0

3 years ago

Why is it very difficult for astronomers to find objects like planets and asteroids?

mezya [45]

It is difficult for astronomers to find object like planets and asteroids because it takes a lot of time to verify the objects locations and what surrounds a certain object in order to prove and be precise of its location

3 0

3 years ago

A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?

VLD [36.1K]

The initial kinetic energy of the car is
$E_1 = \frac{1}{2}mv_1^2 = 1.9 \cdot 10^3 J$

Then, the velocity of the car is decreased by half: $v_2 = \frac{v_1}{2}$
so, the new kinetic energy is
$E_2 = \frac{1}{2}mv_2 ^2 = \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}$
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
$E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J$

5 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

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

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$