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4 years ago

7

Are graded receptor potentials always depolarizing? Do graded receptor potentials always make it easier to induce action potenti

al?

1 answer:

Oksana_A [137]4 years ago

7 0

Answer: Yes,graded receptor potential always depolarize.

Yes,graded receptor potentials must occur to depolarize the neutrons to threshold before action potentials can occur.

Explanation:

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If spring has a spring constant of 500 N/m and is stretched .50 meters,how much energy is stored in the spring ((show work for f

Free_Kalibri [48]

Answer:

The energy stored is: 62.5 Joules

Explanation:

Given

$k = 500N/m$ --- spring constant

$x = 0.5m$ --- stretch

Required

The amount of energy

This is calculated as:

$U = \frac{1}{2} kx^2$

$U = \frac{1}{2} * 500N/m * (0.5m)^2$

$U = 250N/m * (0.5m)^2$

$U = 62.5\ J$

7 0

3 years ago

a stationary police officer directs radio waves emitted by a radar gun at a vehicle toward the officer. Compared to the emitted

Arisa [49]

Answer:

D) Higher Frequency

Explanation:

Higher frequency because it says a vehicle “towards” the officer

3 0

3 years ago

Restless tectonic plates move (shift) between one and fifteen centimeters per __________. 1)year 2)month 3)day 4)minute

Alina [70]

1(year is the answer

4 0

3 years ago

Read 2 more answers

Como anticipan la caída del proyectil, con movimiento parabólico para que el misil no haga daño

Elena L [17]

Answer:

Conociendo la velocidad inicial del proyectil y el angulo de lanzamiento con respecto ala horizontal.

Explanation:

Para poder anticipar la caída del proyectil es importante conocer la velocidad inicial del proyectil y el angulo de disparo del proyectil con respecto a la horizontal.

A continuación se presenta un diagrama o esquema donde se pueden ver estas variables y se explicaran a la brevedad:

Para poder encontrar el rango que es la máxima distancia horizontal recorrida por el proyectil debemos utilizar la siguiente ecuación:

$x=(v_{o})_{x} *t\\where:\\(v_{o})_{x} = velocidad inicial x-component [m/s]\\t= time [s]$

Para poder encontrar el tiempo debemos utilizar la siguiente ecuación:

$y=(v_{y} )_{o}*t-0.5*g*t^{2} \\donde:\\(v_{y} )_{o}= velocidad inicial componente y [m/s]\\g = gravity = 9.81 [m/s^2]\\t = time [s]$

En la anterior ecuación, igualamos y = 0, ya que cuando el proyectil cae al suelo la distancia vertical es cero. De esta manera podemos encontrar el tiempo t, ya que conocemos la velocidad inicial del proyectil en la componente y.

Seguidamente reemplazamos t en la primera ecuacion y encontramos la distancia x o el rango.

8 0

3 years ago

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

$V_1=V_2=50912\ V$

The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

6 0

4 years ago