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mars1129 [50]
3 years ago
15

What structure is located at the front edge of the retina and has a tooth like appearance?

Physics
1 answer:
pogonyaev3 years ago
3 0

Ora serrate is located at the front edge of the retina and has a tooth like appearance.

Our eye has many portions and parts, all of them are there for some function. The ora serrata is the most anterior extent of the retina or serrated junction between the retina and the ciliary body. The retina ends at the ora serrata, where the ciliary body begins.

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En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia
sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

  • Velocidad = 50 m/s
  • Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

Sustituyendo en la fórmula, tenemos;

Frequencia = \frac {50}{40}

<em>Frequencia = 1.25 Hz</em>

7 0
3 years ago
A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr
natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

          f ’= f₀ (v + v₀ / v- v_{s})

let's reduce the magnitude to the SI system

            v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

            v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

          f ‘= 100 (330+ 33.33 / 330-44.44)

          f ’= 97.0 Hz

4 0
3 years ago
A woman throws a javelin 35 mph at an angle 30 degrees from the ground. Neglecting wind resistance or the height the javelin thr
Anna71 [15]

Answer:

35 mph

Explanation:

The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.

When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.

When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.

So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.

5 0
3 years ago
Which term refers to a variable that a scientist adjusts during an experiment
vladimir1956 [14]

Responding variable

Explanation:

7 0
3 years ago
Read 2 more answers
Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m
Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0
3 years ago
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