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3 years ago

Where does an electron in a metal get enough energy to escape from its atom as hypothesized in the photoelectric

2 answers:

8 0

When energy from incident photons is transferred to the electrons, an electron in a metal get enough energy to escape from its atom as hypothesized in the photoelectric effect.

Answer: Option D

<u>Explanation:</u>

At the point, when the light incident on a metal, electrons can be shot out from the outside of the metal in a wonder known as the photoelectric effect. This procedure is additionally frequently alluded to as photo emission, and the electrons that are shot out from the metal are called photo electrons. Regarding their conduct and their properties, photo electrons are the same as different electrons.

The prefix, photograph, basically discloses to us that the electrons have been shot out from a metal surface by incident light. In the light's wave model, Scientists predicted that expanding light amplitude increases the kinetic energy of radiated photo electrons while the increase in the intensity would increase in the estimated current.

Elina [12.6K]3 years ago

5 0

Answer:

D. Energy from incident photons is transferred to the electrons.

Explanation:

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Which one of the following acid-base imbalances? Arterial blood gas levels are obtained from the client. If the client’s results

nekit [7.7K]

Answer:

Explanation:

The correct answer is Metabolic alkalosis (D). A pH of 7.48 shows slight alkalinity, this normal concentration of Co2 in the blood ranges from 35 mmHg (millimetre Mercury) to 45 mmHg and the normal HCo3 ( Hydrogen trioxo carbonate ion) concentration ranges from 22mEq/L to 26mEq/L.

Therefor the patients pH level is high the Co2 level is normal and the HCo3 level is high. Hence, Metabolic alkalosis

7 0

3 years ago

A particle moves along the x axis from the origin. The magnitude of the position vector at time t is

Maurinko [17]

1) The average velocity is $-2.1\cdot 10^5 m/s$

2) The instantaneous velocity is $64t-260t^3$

Explanation:

The average velocity of an object is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

$x(t) = 32t^2 - 65t^4$

where t is the time.

The position of the particle at time t = 6 sec is

$x(6) = 32(6)^2 - 65(6)^4=-83,088 m$

While the position at time t = 12 sec is

$x(12)=32(12)^2-65(12)^4=-1,343,232 m$

So, the displacement is

$d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m$

And therefore the average velocity is

$v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s$

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

$x(t) = 32t^2 - 65t^4$

By differentiating with respect to t, we find the velocity vector:

$v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3$

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

brainly.com/question/5248528

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6 0

3 years ago

85POINTS ASAP!!!!!!!!!!!!!!

alexgriva [62]

The andwer of tye question is 3O2

6 0

3 years ago

Read 2 more answers

Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de

vovangra [49]

Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

donde v0 es la velocidad inicial del auto = 28.7m/s

v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

t = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

6 0

3 years ago

What part of the visible light spectrum produces the most light?

fomenos

Answer: all colors

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7 0

3 years ago