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stepladder [879]

3 years ago

12

For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency

(1.2×1010 Hz ) and wavelength (0.026 m ). Recall that the Avogadro constant is 6.022×1023 mol−1.

1 answer:

Elanso [62]3 years ago

4 0

Answer:

4.79 J

Explanation:

The energy of a single photon is given by

$E=hf$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

f is the frequency of the photon

Here we have

$f=1.2\cdot 10^{10} Hz$

so the energy of one photon is

$E_1=(6.63\cdot 10^{-34})(1.2\cdot 10^{10})=7.96\cdot 10^{-24} J$

Here we have 1 mol of photons, which contains

$N=6.022\cdot 10^{23}$ photons (Avogadro number). So, the total energy of this mole of photons is:

$E=NE_1 = (6.022\cdot 10^{23})(7.96\cdot 10^{-24})=4.79 J$

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You get up in the morning, get dressed, eat breakfast, walk to the bus stop, and ride to school. List three different energy tra

satela [25.4K]

1. mechanical to chemical
2.potential to machanical
3.motion to potintial

3 0

3 years ago

Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the tex

OlgaM077 [116]

Answer:

L = 0.109 H

Explanation:

Given that,

Number of loops in the solenoid, N = 1500

Radius of the wire, r = 4 cm = 0.04 m

Length of the rod, l = 13 cm = 0.13 m

To find,

Self inductance in the solenoid

Solution,

The expression for the self inductance of the solenoid is given by :

$L=\dfrac{\mu_o N^2 A}{l}$

$L=\dfrac{4\pi \times 10^{-7}\times (1500)^2\times \pi (0.04)^2}{0.13}$

L = 0.109 H

So, the self inductance of the solenoid is 0.109 henries.

3 0

3 years ago

The indices of refraction for violet light (λ = 400 nm) and red light (λ = 700 nm) in diamond are 2.46 and 2.41, respectively. A

JulijaS [17]

Answer:

0.42°

Explanation:

Using Snell's law of refraction which states that the ratio of the angle of sin of incidence to angle of sine of refraction is equal to a constant for a given pair of media. Mathematically,

Sin(i)/sin(r) = n

n is the refractive index of the medium

FOR VIOLET LIGHT:

n = 2.46

i = 51°

r = ?

To get r, we will use the Snell's law formula.

2.46 = sin51°/sinr

Sinr = sin51°/2.46

Sinr = 0.316

r = sin^-1(0.316)

rv = 18.42°

FOR RED LIGHT:

n = 2.41

i = 51°

r = ?

To get r, we will use the Snell's law formula.

2.41 = sin51°/sinr

Sinr = sin51°/2.41

Sinr = 0.323

r = sin^-1(0.323)

rd = 18.84°

The angular separation between these two colors of light in the refracted ray will be the difference between there angle of refraction.

Angular separation = rd - rv

= 18.84° - 18.42°

= 0.42°

6 0

3 years ago

A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard

podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0

3 years ago